Scribd
Upload
180 views142 pages

Maths 2b LM

The document contains a series of long answer questions (LAQs) related to circles, including finding equations of circles passing through given points, proving concyclic points, and determining tangents. It also includes practice questions and solutions for various problems involving circles and tangents. Additionally, it covers topics such as chords, areas, and differential equations.

Uploaded by

koushii.2k
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
180 views142 pages

Maths 2b LM

The document contains a series of long answer questions (LAQs) related to circles, including finding equations of circles passing through given points, proving concyclic points, and determining tangents. It also includes practice questions and solutions for various problems involving circles and tangents. Additionally, it covers topics such as chords, areas, and differential equations.

Uploaded by

koushii.2k
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
You are on page 1/ 142
LONG ANSWER QUESTIONS (LAQs) T ws japtor 4 (Circle) QI. Find the equation of circle passing through three points. (3,4), (3, 2), (1, 4) march-23, a18 | Aua-22, 028 | May-t8 18 | March-16, 018 (ans : Page No. 18) Practise Question | 7. Find the equotion of ccle possing through each of the following three points. (12) (3.= 4). 5.6) Benet ee May-8, 018 | March-16, 5 jing the @2._ Find the equation of a circle which passes through (2, ~3) and (-4, 5) one eae centre on 4x + 3y + 1 = 0, June-23, a18|March-20, 218 | June-t9, Q18 | May-17,a18 (A Practise Question ing ei PQ. Find the equation of a circle which passes through (4,1) and (6, 5} and having the cenlte 6h 0 3y—24=0. [ANS:x?+ y?~ 6x-By + 15= 0] Zand deduce the value ay 18,022 March-18, 016 (ans: Page No.4) ‘cot"xdx.n being a positive integer, n 2 2 and’ 1, Obtoin the reduction formula for, = f (ans Page No deduce the value off cot*xcx may-24,az2|Marse-12, a2 (022. Obtain reduction formula f tan’x.dx for integern > 2 and evaluate, franca cee June-23, 022 | May-17, 022 @33. Oblain the reduction formula for!n = Jocos'x ax, nbeing a positive integern > 2, at (Ans Page! deduce the value of | cos’x dx June9,a22 being a postive integer n> 2and hence: en Practise Question +. oblin he ection ora for trove ot xcs Ma, 022 wear 22) fede Jorn ssc 3, ae IANS 1,= 34. Evaluate fe*sin(bx +c}cx; (0, be eR,b#O)onR. Chapter 7 (Definite Integrals) may 22, 098 1AQ36, Evaluate feo som 23 | 1aaa7. ele +O ; 3 F eine 208% gy synons.3 mer taase. Evaluate | 54 165in2* 3 fova% srch-24, O23 | as. etc free : | [racine eon es x sink eoosT of toa ws: = _ (pase [ee met hs PG | sh | LA@AO, Evaluate yo fac=o0b=9 GND =W).dX Mareh40, 24 13,23 | March-15, 23 XIN Gy. say 1AQ41. Evaluate f Tooke MY sune-19, 223 LA@42, Evaluate feizis stay 24,023 Aug-22, 86 | Mare 20,073 | ne $9 (ay -17, 029 LAQ43, Find: J xin? x cos! x cx. March-19, 023 2 LAQ@44. Show that the area of the region bounded by * 2 pe (ellipse) is rab, the area of the circle x? + y?= a 2 : LA@45. Show that the area enclosed between the curves ¥°=12(x+3) and y?= 20(5—x) is 65/2 Chapter 8 (0ifferential Equations) LAQMé. Solve the differential equation (1 + x2) $% + y = tan-'x may-24, 026 LAAT. Solve the diferent equation (x + ¥) Es DxyGX marceasere sq.units.. | Practise Question | | PQ. Solve (x? + y'dx = 2xydy. (ANS. Treas ones I FQ. Soive xe («-2) a psy Lav \Q50. Solve the differential Al equation 2 Practise Question PQ. Solve (x+y + 3) R QUESTIONS (saqs) 4 (Circle) Find the length of the chord formed by xry= cron the line xcos.a.+y sin o.= p, marcas, ars (Ana: Page Wo. 89) Find the length of the chord intercepted by the circle, x°+ y2—x + 3y-22= 0. on the ine Yo x — 3: Aug G17 May-22 028 | Mare-28 ONY (Ans : Page No. 59) Find the length of the chord intercepted by the clicle, x'+ ¥—8x-2y-8 = 0 on the ine xy +1 0.smeah at [Macht att eee Practise Question “pa. nd the length of the chord intercepted by he cicle x’ +/+ 8k-4y= 16 =0.0n the ine S.-Y + 4=0, (ws: 20761 Find the pole of 3x + 4y- 45 = 0 with respect fo the x? + y*~ 6x — By + 5 = 0. may22,ar7 (Ans : Page No. 60) ‘Practise Question ra. “Prine pole olx+y+2=0winrospect 1 the role x + y2— Ax + by Find the area of the triangle formed by the normal at (3, ~4) to the circle x? + y?-22x— 4y +25 = Owith the coordinate axes. March-18, 011 {Ans : Page No. 61) The liney=mx+c and the circle x°+ y*= aFintersect at A and B. f AB= 22, thenshow that c2= (1 +m?) (a?— 22). Aug-22, 29 March-19, 012 (Ans : Page No. 61) 28) Find the value of ‘k’ ifx + y5 = 0, 2x + ky-8 =O are conjugate with respect to the circle x* © yi 2x-2y-1=0. May-18, 011 (ans : Page No. 62) Find the equation of the tangent at the point 30° (Parametric value of 8) of the circle x24 y2+ 4x + by 39 = 0. Jne-19, a1 (Ans Page No. 62) | Practise Question nolege | P@. Find the equation of the tangents to the circle x? + y?~ 4x= 6y +3= +142V5=0) which makes an congle 45° with X- Find the.equation off the tangent to x? + y? - 2x + Ay= Oat (3, -1). Also find the equation of tangent parallel to it. may-17,a11 Page No.6 . Ifa point P is moving such that the lengths of tangents drawn from x2+y?— dx— 6y- 12= 0. and x? + y? + 6x + 18y + 26 = O are in the ratio equation of the locus of P. areh-17, a11 xis. [ANS :x | Practise Question ae | PQ. If a point Pis moving such that the lengths of tangents drawn from x? + ¥? + Bx+ 12y + 15=O.and x? + y?— 4x 6y—12= are equal then fing thy TANS 4x + 8y + ne? - Find the equation of a circle whose center ies on and (4, 5) march.15, a11 {9,2} 101ne ciclo x + yo. from nd Year) yents ow” welt renee re ere Dareet ery &- 89472010 4 a Feaneminerset 1,2) of he Eriact. ‘ SA@I3, Show that the rongent ah its point of hs wea yet det Y= 5 ; e jrcle- -2y= 2 is 8 and dista Chapter 2 sptenot Crh! jojowind MO, ayy? 6x -2Y =O: Mayan Grate na ine raceat cone OMS 057 gy ea ee xe tytn axn yt 22, 20 | May-18, O12 the otipose * 121400 weston ye town 1 6. Find the ea a. Fane ocscl cor ve aye eax SY-P™ ‘June-23, 214 | pp syte ae-72 0.20 ih of the com! a7. Findthe ec sion and lengtt 3 SAIS. Dd the ed oe jzoandsi= et Y ; . of the folo Sexity?+3xt Sy mon chord of the following circ} May-A5, O43 [5 ‘com SA@16. Find the equation and length of he x ty?t xt 2y+1=0 Find the te of centre, yee yt dct By +2= 0. Marcht7 a2 ; ° : : . common chord of the following circles. rain = ale ey pans :x#2y-2=08 2/81 eae ee aa + y= Sx~ by + 420.74 2 aa a re i i th the points of intersection of SAQI7. Find the equation of the circle passing throus} , “a ‘Coon xe ty? 8x- by +21 = 0: xitye—2x-1. ‘and (1, 2). May-7, 12 ‘ SAQI8. Find the equation of the circle which passes through the points (2.0), (0,2) anda to the circle 2x? + 2y? + 5x- dy +4 =O June-t2, a12 aa and having the centre at (2, 3). Mareh-25, a12| May-22, Q19 | March-19, Q14 _ ‘$AQ20. Find the equation of the circle which passes through the origin and intersects the below, orthogonally i x?+ y?— dx + by + 10=0, X24 y? + 12y + 6 = 0. June-23, 012 | Aug.-22, 019. Practise Question : PQ. Find the equation of the circle which orthogonally H+ y= Ax 6y-3 exis yeday 4 sAQai. el pe euation ofthe circle passing through the origin, having its centre on he ne specing the ciclex’+ y2— 4x+2y+4=0 orthogonally. augsahio [ay 2.0% TANS =32 + y+ 6x — ay = 0} SAQ22. Show that the circles x2 + other and find the Point of Practise Question PQ. Show that the circles s Find thet point of cont Y~8x—2y 48 CONtaCt, May.22, a20 B24 F—2x— by 20 = 9, § lOct. IANS : 5,5) SAQ23. Show th : lat the angle betwee, een the Circles +y2= — y @ inthe standard form such that eistance between foc! re heen drectices B22: erat! May 17,044 (ane: Page No. 72) nce jon tor the line +MY * tor 4 lve ion ofthe ose 101 sage 1n=010 be a tangent to is. Find the condi Se Yar manana the ellipse find the equation of tangent ‘and nor (nn Page No. 72) mal to the ellipse x? + By? = 33 at (-1,2). (Ana: Page No.72) dune, 014| arch-20 244 notesot ocl lengthoflatusrecium and equationsot directrices ; panacea a n 1gy— 36x # 92y — 92 = 0 Maen, 3 | Maret a4 | Mey 8 a4 othe eee Gocteaee noon - axis, minor ox’, latus rectum, eccentricity, coordinates ions of directrices of the olipse : 9x? + 1éy"= 144 aye ata may-7-N4|March-10,<13 (Ans: Page No. 73) Find the length of the major of centre, foci and the equal Fopaarch-24,Q13 | Aug-22, 22] March 20,213 | oe practise Question Te pho merc cnis minora avs eclum. eceoniiehy,eoersnetea tcenalad + Fin he equations of directrices of the following elipse. 4x: + y—@x*2y +1=0 ie «4 Length of the minor axis = 2, Length ofthe fats rectum = 1, Eecenticy = 4 ps Lana te mao : 1, ‘Foci = (1, =f #3). Equations of Directrices = V3 y+ v5 #4 =O] Coordinates of Centre = 1 red to its major and minor axes as the coordinates find the equation of the ellipse refer ie oe i ith latus rectum of length 4 and distance between foci 4/2. gxes X Yespectively with latus rectum 1a pen foci 4/2. Find the equation of the ellipse with focus at (1, -1),e = 2/3 and directrix asx + y * 2= 0. Mareht9, a13 (ns : Page No. 74) ‘and the length of latus rectum is +>. aug-22, zt May-22,22| March-5,013 (Ans : Page No.75) Find the equations of tangent and normal to the ellipse 2x? + 3y = 11 at the point whose ordinate is 1. March-19, 214 March-16, 014 Obey (Ans : Page No. 75) mit OF . Ifthe normal at one enc! of a latusrectum of the ellipse > oa = 1 passes through one end of the minor axis, then show that e¢ + e? = 1 [e is the eccentricity of the ellipse] Feb/March-24, Q14 | May-17, Q13 (ans : Page No. 78) 934. Find the eccentricity of the ellipse (in standard form), ifits length of the latus rectumis equal to half ofits major axis, June-23, 013 Practise Question : PQ. If he length ofthe latus rectum is equal to half of find the eccentricity of the elipse. {ANS : “2 48é. Find the equation of the tangents to the ellipse 2x: + y= 8 which ar (i) Parallel to x- 2y-4=0 aad a oy (Second Yeon?) Z ‘ sy ermrtr set eee nypernaia oe ~4Y 12. which j gent ® sacar. Find the equation of 19 ] {y Parallel and we a i aa aT ona arch endicuiert0 he ee (amps i ae neo ee sre rena Ae se rar res tne equations (y Porte! (0 Popendiu ns 227" 0,020 joxto he tne x+ 20 pave) se 3 i \d the length of latus ‘city, foci directrices oO” SAQ38. Find the centre eccenticit. e hyperbola. gx 9y2= BK 32 = 0. Aug, OF equation of the directrices, length of the lots, Find the centre, foci, eccentricity. of the following hyperbolas. fi) Wey? 9xt= 144 (i) x2 4y2= 4 snes, ats Marent®, O15 ; SAQS?. Practise Question Facies austen joc secennicy, equation ofthe ciecices lonoi ol RGI falowing hyperbole. 54+ 2088 anv a=. Eeanty =f Eqn fhe (ean SAQ4O. Prove that the point of it x 3. —4=ieson the cle ¥+y*=!—b* Mave. ats| March gi 3 SAQA. Tangents to the hyperbola 27 ~ br hyperbola. show that fhe Point of intersection of these ta k(x?) when tan 0, + tan 0,=K June-19, 15 Chapter 7 (Definite Integrals) io sin’x SAQA? Evaluate [ps8 A a= dx June 23, 16] Maren, a16 Practise Question PQ. Evaluate f etek fe | ice ET ; SAQAS. Evaluat x : J Xeag Mavs, cas Practise Question ce ra. ft 4 ox {i ANS <4 +09 5) (sin arch-23, 023 | March-20, 96 ‘ aveatatss sea toe re } er, Pooos [ae ere tpt d frie-aicx Load (nm Page Mo.) . Evaluate ef (Vee) ox. rebsmaren24, a8 Mareht8, 8 ince ee _ Find the area of the region enclosed Dy the curves y =e, y=X,x=0,x=1, (Ans : Page No. 84) Febsatarch 26, O16 . Find the area enclosed by the curves y? = 4x, y2=4(4—X) June-19,016 (Ane: Page Wo. 84) Find [ sintxcos'x dx March-18, 018 Garironte PQ. Find re sin?x cos*x dx. IANS: $1 8 (Differential Equations) re _ solve the differential equation So =tan®(x + y).sune2s, a47|Aug-22 037 | May, a8? (ns Page No. . Solve the differential equation oy _xyty ox” xy tx 3} . Solve SY+ 1 = 27. Marne, a7 ay : Solve Sra e™Y+x7erY way24 a7 |Aug-22, 26] Mey22, (02r | May.17, 17 5. Solve (xy? + x}dx + (yx? + y}dy = O Mareh-20, at7 March-23, Q17 | May-22, 026 Practise Question PQ. Solve (ex + I}ydy + (y+ I}dx=0 [ANS :e'= +I) +ee) 16. Solve sin-'(34)= X+Y March-20, 024 7. Solve aw =xtan (y - Solve . Solve Gy + ytanx =Cos*x May-18, 017 9. Solve V1+ dx + Vit y2 dy =0 aug.22.0 Chapter 4 (Cir ent from (2: 5) 46, a2| March-18, Q2 VSAQI.__ If the length et torn 24 on 022 a1 and radius r where find k. mente ot I rthe ccles with Ce i, he equation o| woe Sea. rageeaate pm cee s= Owners Ponda it fhe equation of the norm aes ; VSAQ3. et 5) 3 xbtyh=lOx= 2y + 6 Marchds, gh (2-1) 08 ea a4. 4 a the equation of the circle passing Lo a = centre C= (2,-3) and radius r=4 (2, 3), May22, a1 : VSAQ@S. Find the equation of the circle with : ith radius 6, then find the VSAQ6, ixt+ y?=4e+ dy +e = Orepresents a circle with radius 6, then Fevsmarch-24 01 | Aug-22, 3 | May 22, 2 | May-18, Qt Find the center and radius of the circle x2 + y?— dx By— 41 = 0 March-20, a1 fi ri le 2x? + 2y’ VSAQ8. Write the parametric equation of the circ! 2 & ae VSAQ9.__ If the length of the tangent from (5, 4) to the circle x* + y* + 2ky = Dis 1 VSAQ7. March-20, 02 | May-15, 02 | March-18, Qt VSAQI1. Find the polar of (3, -1) with respect to the 2x? + 2y?= 11. June-23, a2 Junet9, VSAQI2._ Find the valve of 'k' if the points (1, 3) and (2, k) are conjugate points: the circle x*+ y? = 35, May-22, a3 | Marcht9, @2 | May-17, @2| March-16, 2 YSAQIS._ Find the chord of contact of (1, 1) to the circle x2 + y2= 9, May-24, a2 | VSAQ14. Find tt hi cn it eae Chord of contact of (0, 5) with respect to the circle vars YSAQIS. Locate the position of the point P(3, 4) wi 0° 4) with res i SEX ty? 4x 6y-12=0 ‘Aug.-22, a2 : ee ae VSAQI6. Find the equation of the circle eS itcle passing through the. origin and having 44) evatuate [115225 x On (-$:) sam (oe Page ser (1+) R:cos(xe*) = 0} march es evawate | 0,a¢1 ondb>0,bA1)ONR. may-22,039 (Ane: Page no. 168) 1 shal on (+1, 1) May-24,07 | May-22, 09 (Ae Page Mo. to) | Evolvate [Speer onlcR\ [[er-05 e 2} [amet (5):n <2|]araas E (Ans : Page No. 108) | Evolaute J sec?xcosec"xdx onic e\inen e Duf(ans1)5:n ez} une-23, G7 | May 18,08 | \ March (Ans : Page No. 108) ster 7 (Definite Integrals) aT . Find f cos®x cx maren23,a9 Arey 6 Find fhe area of the region enclosed by y=%° +3, y= 0.x: May-24, 018 Marchv2, 09 |May-17, 08 eh 2/2 Find the valve sin! cx Juneas, a8 “2 . Evaulate [ cos!'x dx. may-24,a8 ars, 8, é |. Evaluate : fiarerax iay-24, 08 5 . Evaluate f (x+1)cx ANg-22 018] May-2, a3 é 2 Evaluate [I~ x1 dx avg-22,a14] Maret, cB é Evaluate { y2¥26050 dO June2s,a8|may.22,014 3 i 2 Eval x late ay AX Feb.March-24, 09 | May-18, 8 2 . Evalu aoe e ate i) Taye OX March-23, 28 | March-20, Q8 | March-16, 08 i es the value Of K. Marcha, g aay 5 Fin ‘ ndiculors from any Point on iso yangent tO ne perPe’ Waa toes + k=0F cocaine es iuate vanae. Figsing rower iene eT Bs. Oe oimosrmetle 0, ay, ne the oy vsaa42, tine angle bee find its eccentricity, Evaluate i », 0), the vet ose foci are (#5. ) trans an 3 the | sa@43._ Find the equations of Ine length 8, aay-22, 0231 Mer ae x rola => — totes of ahyperbola ar twoasymp! ingle between the VsAQ@44, Show that a alvate ‘of 25e¢"!(@). March48, 018 Pane ‘yioreca aa SA@45, Find the equation of the normal at @ Chapter 6 (integration) vsa@4s. Evaluate [ 2xsin( sgax on March-23, QT | June-19, 6 ix2-+ 1) dx, x € R Jones, a8 saw, Evobate f= VSAQ48. Evaluate oF “one. ay 22, 12 dx on ICR\{(2nt1)x:n €Z} March-20, a6 Vsaa4?. Evaluate f ee VSAQ5O. Evaluate [sin7xdx xe R auga2.a11 YSAQSI. Evaluate { /I=cos®xdx on|c [2nx,(2n+1)n],n © Z. marcha, 08 WsAQ82. Evaluate f1*E° aoe oS onleR\ (min © 2) March-19, 08 Vs 1953. Evaluate [> wl 7X ON R. March20, a7 x VSAQS4. Evaluate f lane xlogx[log[iogx}y %O0 (1, 00) marchas, a7 VSAQS55. Eval en ‘dx VSAQ56. Evaluat le YSAQS7._ Evaluate fe Gesloge ar OF (0, =). Fen WSAQSB. Evaluate fom WDA [Jingo Tax xo} ®°) Aug.22, May-22, a1 a MC (0.0) augue 2 , VSAQS9. Eval lvate 1 Sey KEE H Onc fop 12 ~) Fe MMMar.26, 06 | Aug.27 Fs, ance on oie vs” protusle fm "9 { xontcxmercst ot uate vsaa7e. Eva! intx COs r see cosx wate vsaa7e. Evauels dx Avge, 074 jons) chapter 8 (iterentil Eat f ysaq@en. Find the order and degree Of | a3) arch28, 91° : of the order and degree VsAQBl._ Find the order o |S oy (¢ dy aly By) | aoe e differential equation may2a degree of th VSA@s2. Find the order and s 2, slay VSA@83,_ Find the order and degree of the differential equation x: March-15, 10 VSAQ84, VSAQBS. VSAQ86. gy a a et 4+ x25 ca ey+ey YSAQ87._ Find the general solution of x + y a YSAQEE,_ Fhe gener soition of Y= oy want, Va VSAQ89._ Find the general solution ot Ey dx — yx March-19, Q VSAQ90. Solve WY l+y? Te mesh ate VSAQ9), Fotm the differential €qUation cores ITESpo} CIRCLE the equation of circle passing through three points. 4), (3.2). 04) March-23, Aug-22, May-18, March-18 ints P(3, 4), Q, 2), R(L, 4) be, ofthe circle passing through the Pol fl) equation pay tgs tay ten 0 Equation (1) is passing through P, 4) r+ P+ 280) +278) += 0 9116+ 6g +8 en 6g + 8ft c= -25 Bguation (1) is passing through Q@,2) @?+ QP +28@) + 72) + 6-0 g+4+ 6g t4fec=0 6g +4ftc=-13 tion (1) is passing through Ry (a? + (4)? +29) + 24) #0= 0 1+16+2g+8f+c~0 2g +8f+c=-17 @)-2) 66+ 4fte=-13 if = 2 f=-3 @-@) 6g + 8fte=-25 2g + ft e= 17 4g 8 => g=-2 ibstituting g, f values in equation (2), 6(-2)+8(-3) +c=- i e=11 oi ; Riana oat Substituting the values of g, i i a ty + 2-2)x+ ae me "x + - 4x — 6y + 11 = Dis the required equation of the circl 4) “LAG2. Find the equation ay Ranta cakesy st asce Let the equation of the circle Ber seep tage tay ten The point (2,-3) ties on the circle i.e eye (3)+ 2602) +309) 479 = 449tdg-ofte=0 i => 4g-etcr13=0 Another point (-4 5) lies on the = CaPrer+ 2g) +770) te" = 16+25-8g +10f+e=0 => -Bg+10f+c+41=0 ‘The circle is having the center on the line 4x * SA Ce ore = -tg-3f+1=0 From equations (2) and (3), 4g-6fte=-13 pale sig a285" 12g - 16f-28=0 3g -4f-7=0 Solving equations (8) & (5) circle By +1 > 3 T6+9 25 eoaran! -2 8357 fn oe Substituting the values of g fin equation @), 4()-6(-1) +c+13=0 c=-23 S=1f= 1 oes Substitutin, g the correspond " ponding values i The requis fe : Tequired equation of circle is Fi in equation P+ P+ 2AM)xe eyes ee by, Bt +2e-2y 23.00 Ee Wea 3228 = =Oie, center = (-g,-f) ou oR re fe WY) : eo o & e Q, lamin denne os ee yt get ugh (1.2) i jation Since 2g) +270) *-= 9 aged ae 6,0) ‘equation (2) passes through ¢ ince Gy + + 2gC-8) #370) * = 9 Ang + ¢= -36 on : tion (1) passes through (~ Sines fa 2+ 26-2) + 270) +69 4g + af c= -8 -@ = , © Gaerignen 2426 dg + 2f= 34 gt fei -@ = Ag +c+4g-4f-6= 36 +8 8g - Af = -28 5g-10 = g=2) ubstituting’s’ values in equation (6), Substituting g, f values in equation 2), 2Q)+2()+C= 2 2 Substituting g,f& c value sin can (1), xP+y?+2(2)x+2(3)y-12= 0 Substituting (2 -8) in above equation (2+ (8 +402) + 6-8) - 12 =4+64-8-48-12 = 68-68 =0 Given points are concyclic Let the equation of the circle lie Suxty'+2ex+2fytc=0 Follow same procedure as above. (6,10) lies on the above circle. 3) and (0: © OO se. yatta? Je be, oto ts on the cle : meee 0) and le $= {a Dlieson the cine TO p+ 200) +S ae 6-0, > ba The point S ztee= cirel 7 A ircle The point (45) Hes on es (5) sete 0 at GF 42800) gM = ~@ > griofrer- @-O 5, 107+¢2-41 aca Rages => 4g-2/+3-0 ..@) HIM Solving equations Pp and (6), a 1 6 Substituting the corresponding values in equation (2), =13 fe )ten—4 Senge Fd 3 equation (1), UUuUUY ee -13 3 eel. Heo *¥) - Ir ae ae +1) #44 sgh ited equation of ies ue = u Se The pery to the line 3x +4ys4e + 2R+ y-3F Peat tye P+ yt dx — 6y ’ BHP dx — by + equation of the ommon tong” ~ 100 « dane ne ay + 100 #0 aay 18, mares circles are, ete yt 22x Ay 1000 axe y- meray 1000 cG=(-11,2) = is 4100 = 225 =>, =15 ircle, $= 0, c-(1,2) = (125-100 = 4=5 2M cc iy 2-C 2a = (C2 +@? = (484716 = 500 C= 10/5, 1#1,=15+5=20 WC,C,> 1,41, 0 circles are away from each other Also, r,:1,=15:5=3:1 Cin the ratio 3:1 externally. p= [3Q0D=MEn) 3€-2)-12) 3- 3-1 E=(22,-4) tm be the slope of a common tangent mal center of similitude ‘E’ divides 2M Bquation of the direct common tangent is wen a g eye mex) yt dye me 22y i mx -y=22m=4=0- A) 5 ecquation (1) is tangent to circle S,=0 then, red ; d is the perpendicular distance from. Ue (11, =2) to the line mx = y= 22m=4= 0 [Lim +2=22m—41_ thy te me ae Mm -21=5¥ne 1 Squaring on both sides, 121m?-+ 44m +4= 25m? +25, => 96m? + 44m—21=0 ieee 2(96) + 41936 + 8064 192 oe 44 £100 192 loa 444100 == 1007 192; te arlozae = nfo ee Case (i) If m= 3p then, 7 (y+4)= ag 2) => Pay +96=7x- 154 => 7x-2Ay-250=0 Case (ii) If m= I LAG. ‘Find the transverse common tangents of the clicles 3 xi + yl dx 10y +28 0.and abt yt dx by +420 Junot Marent7 Sol. Given equation of circles are, S,axtt y- 4x -10y + 28=0 Spextyp+dr-6y+4=0 For circle S,= 0, G=25) 10 VaF25-28 Substituting the correspongin a equation (1), 4 s = For circle $, = 0, Case (i): is aan : = (23) y-$= a = VOPR FCF 2y-9_~3 - a = See “peep > on=3 22M = Ay-18=-3x43 The internal centre of similitude () , => 3xr+4y-21=0 divides C, C, in the rato r,:r,= 1:3 internally. 1 Ifm=co= 4 then, wee 430) oF Y-Z=g@-D El => x-1=0 = (ye) 1m | *- Equationsof transverse common are, 3x + 4y—2] = -1= 2 bem, tthe slope of transverse common tangent 4-290 Equation of tangent is given as, ¥~Y=m(x—x,) 9 = y-Fema-p (1) > 2y-9=2m(x-1) => 2y-9=2mx-2m => 2mx-2y-2m+9=0 ) Equation (2) is a tangent to circle. S,=0 Sone is ofcitcleis the perpendicular distance from centre 2,5) toline ie, equation Q), = we : nnaK-1 txt tytn dae H 41980, x+y" Show Mat ealne equation of common Ton wyen equations of elreles are, vit ynGr~ Oy F190 9 For citele, $,= 0 ara nav 1 +(8h 0 = VIF OF VG Consider, CC, = 9a8 G- + (5 onsider, |7,~ From equations (1) and (2) CG In-nl ‘The circles touch each other internally. The point he ratio, VB. e5= 1:2 Trae Pio | aS m=-n? m=n P (x,y) = (5,1) ‘The common tangent at this p $,-S,0 3 = A+ 7-6-4 Boe => -4r+7y+13=0=34x- 1 equation of common ta uch each other. Find the rary & “1 point of contact, 5 rah, March, May, Morena fons of circles are Given er gent #1= 8 freeg ye? For circle $= ; Gea "tas Arc 7 Eeres poredrele $,70 G=(14) ef a ne OCOD = ier9 = V25 = OG,75 = 4 tye3+295 From equations (1) and (2), CGnn+7=5. The circles touch each other externally. ‘The point of contact P (x, y,) divides C, C, in the ratio r, : Fi = (come mt) 1,=3:2 internally,” mn’ ms: (ceve2e) (34) + 2¢ 3+2 3 = Py) = +2, Bi oe 5 P(xyy,) = Gz The common tangent at the point of contact is, $,-5,=0 = fey Or 2y+1- 2- 2x + By 1329 = -8r+6y-12=0 > 4r-3yt ‘The equation of common tangent at the point of ¢ yen) 21) FER Fea=4] ua 5- SOHO at) [ewe 143) @> ea-(* hei=W = h=9 k+2-10 = k=8 Centre C= G5) “The equation of the required circle is e-mpt yee (e-9F + W-FF x-18r+81+ yi Ioy + 64-25 18r—16y+ 120=0) Rae ep eG : xi yh 908-27 OT anos of £ | find the internal and "Sol. Given, | “Equations of circles, | spare pote ty +990 = os s.exe yp + 908-29 410 1a For $,=0, | q-@-3) ay = 5-5 | 1, = Vay+ 3-33 * (949-33 1 For $,=0, =C51) [easyatiyal = Vz05 41a = 1225 = 18 ; = fCisyp+ (1p 1 = ¥22541-1 = S aa To find a ee ae tangents, that exists for pair of ore 0 nu Gerth Ino nl GG = VO-Cis SI GG = ¥500 = 10/5 1,25,7,=15 rr 5+15=20= 400 ‘Comparing values of equations (3) and (4), >rtr 1AQT1. nin =5:1521:3=m:0 The internal centre of similitude (ic a Sar ee) = (1615) +37) 1(1)+3(-3) P| TT TSR ) -(Bt2) (8) > Internal centre of simititude = [3, The external centre of similitude Q = (Mans m= nay m-n * m—y 15)-37) 1(1)-3(-3) ay ous) iz =a) = (18, -5) Extemal centre of similitude = (18, -5) ‘Show that the circles Kiet 1229 at and etch y)— OR 47-327 Tipe ond fd ter pow of contact: ets “The given circles are, ie pt yendea 6y- 1290 5a + y)-8r- eee se Ey fae From equation (1) qe) yn VERSIE = VETTE = 1B =S From equation (2), 47 o=(% 8) : OTe [TBH _ (25 _ yo = mM -/@-@2V™ a . [FB - (5-3 Distance between the centres is, js ae Ve-3}+b-3) a In-nl = 5-31= 6G=In-nl @2M “The circels touch each other internally and the point P divides in the ratio 5 :3 externally. ola ces m-n san, ae n hes 5(4) 2) 5. ie (ia) i pee gee mee 0. x(- 1) + y(- 1) ~ 2(¢- 1) - 3(y-1) - 1: -x-y~2x+2-3y+3-12=0 =0 -3x-4y- 3x4 4y+7=0 ‘The point of contact is (- 1, - 1) and the required equation of on of circle I Given equatl Aw emt ye? Pop yh) Pome? alee “The equation of Pai $7 25'S angentsis given Ce a 4 oil 2 Zy+3) se att) +y@)- 3+) * 2 j -11 |= gaxtsy-r- 12946 b> Se-8 | = 5,-a%+@F-20)+4@)- 1 ‘ > S=o Sei pte ceercing values in equation) a > By-6f=9 [s+ y= 2x + dy 11] => By- 6 =9x2+ 94° - 18x + 36y - 99 => 25y? + 36 - 60y = 9x? + 9y? - 18x + 36y - 99 => 9x°- 16y?— 181 + 96y - 135 = O's the required equation of pair Angle made by the tangents to the circle is given as, la+6l Voa~6F+QiF Herea=9,b=16,2h=0 [9-16 V9+ 167402 7| * Vas cost = cos 8 = oN — Ron % : vn Eh 1g in standard form, “jue-28 March, A022, May-22, MareN-20 say40, March, March, March-18, rojection of ‘5’ on a shown in the figure | pe the midpoint of SZ and YAY be the straight line through A, parallel to the \ pase 4 1 tne equation of parabol _zbe the focus, directrix and P! pexXaxis and YF be Y-axis, => A is the: origin 0) a2m | 0), (a> 0) then Z = &a,0) and the equation of directrix ++ 4-0 ure, if P(e 9) is point on parabola and PM is the perpendicular distance from P to v P) = (PMY? < ap ty= (eta? - caiving equations () and 3a+b=3 satb=-2 ae The axis 3(Z) +63 Xan bola is | canas Then the equation of Par @ 9M . pene BaD eo xeay thy te ieee 3 am Bote Parabola passes See Sibel ‘The equ yyc=(te zie y An(2, a int b= % inequal , rol the 2 Trequation (1) passes through the Pow! 4 sy A= (2,1) ther, 5)+(B)+e=2 ‘Since -aeatiy +b) te Z a > -2aatbee = ca-24+5-9 > Om v reat. fee pce Substituting the corres equation (1) passes through the PoIMt | eayation (1) ee a B=(1,2)then, x= (Q)v+ (By-t0 on 1=aQ)*+0Q) “oa > 1n4a+2b+e 20> oe Squ 2. 4a+2b+c=1 @) IM => 2v=-5y + 21y-20 as If equation (1) passes through the point | 140 required equation of the = C= (41,3) then, Sy? +2x-2ly+20=0 ee eee: = => -1=%+3b+e > 9a + 36+ = -1 21M a Ss Solving equations (2) and (3), atbte=-2 4a+2b+e=1 =3a-b=~3 = Batb=3 (6) Solving equations (3) and (4), 4a+2b+e=1 9a+3b+e=-1 Se Sa ~b=2 > Sitb=2 6) | Mathematies(B) (Second Year) savation ot ne common tangents the cicle x+y" «2a andthe etore yen 220) aon Giver. Equation of the circle is, aepeae Here, P= 2a uation ofthe parabola it ye tar = () Equation (1 sin the form of yn der = ane “The equation of the tangent to the parabola is, yomrs = Seman ® zm ‘Since equation (2) is tangent to the circl xi yes 2atthen r= d 0+ 22| [24 rates me} 2 Gadmi=|72| swe get, ‘Squaring on bot > meets squaring on both sides, 2 (1+) = 5 = > => mt +2? mt-2=0 > > 4 me (1+ me) =2 nett =2 me (ni (n?-1) (n? +2)=0 me+2=0 => m=-2 => m= 2 [’¢ Roots are imaginary] me +2#0 m-1=0 = mal > madi > m=#1 ‘Thus, the required equation of tangents are, yom+ 28 [- From equation (2)] 2a > yr thts => y=(el) (x+20) yat(x+2a) 1aai7. Sol. ees sq. units were YoY? (y-yl Given parabola is, = 40 Consider a APOR with the vertices as P= (at?, 2at) = Cy. y) Q= (at, 2at,) = Hy Yd) Re (at, Dat) = Cy ¥ ppore a fat Te eat 2|2ar,- 2at, 2at— = oleae) -a 2 ats — ats 2a, (6-)2at-9)| 2 Eat Kit em 81-28 Hota ar = Bia averse BI = (4-4 \—oM4-4)| , = e-ayo-ayla-4)| Multiplying and oe with 2 Area of APQR = 55. 2\G, =1)(t,-4)(-1)| - at eee cee iy “ |2a(t, ~1,).2a(t, ~1,).2a(t,~1)| = £1ea,- |(2at,—2ar,)(2ar, — 2at,)2ar,~ 2a) Area of = e AF Of APQR = 3 1(01— ya)(¥2 2Y%s-¥)I square : units. eee ‘drawn to the parabola y'= Aax ond there ; xtemal point P tangents oF anger ss, such that cot 0, + cot 0,18 a. constant ‘Then Maronty from an © horizontal tangent Snow that all such Even equation of the parabola is, cot 0, + cot 0, = 4 fon of any tangent to the parabol is, yam the tangent passes through P GW) om mex ta Vico mye mex te = mix,-my, +470 “Equation @ is in the form of quadratic equation ax! * bx + 6= the sum of the roots is, : eee =x coefficient - Sum of roots = <3 scefficient | coefficient mym= % ere mm, ae marth slopes the angen From equation (1), $3 reGRATION acosxt S8IDX dx “aan tvatuate { Heose? 59* | Soh } Given integral iS | oosx + 3sinx nae [ESS six * ‘This can be written a5, mminator) | : leno! ay ton + Ba : Numer ae y + B (4 cosx + 5 sins) { Bee ounce ae Woot oere | oy fatcaian)+s(ous)*B COSTS EES Deere Berg ase CAs sB yor (AAS Eee ‘Comparing the coefficients of sinx and cos, , 44 +5353 a 5A+4B=2 a 5x(1) > -20A+25B=15 4x) > 20A+16B=8 41B = 23 3 > Bsa st 23 Substituting B = 2% in equation (1) -4A+5(33) =3 Betigy a ills 4D 2cosx +3sinx 4cosx +5 sinx Sone Vex +16 LGA 20 +10)+B 2x +5 = AQx-2)+B +5 = (2A)x + 2A +B) g the coefficients of x’ on both sides —DA=2=A=1 Substituting the value of A in equation (8) then B =7 a f LF (x? — 2x +10) + B =f sa [+ From equation (1)] 2x -2 ae iM =f Tae aT0. +7) Vea ox¥ 10 oa = 2/PmarFT0 +7 [ee [J FB a= 275 mM Ve ae 7s) = 2/2410 + 7sinh-( 254) +6 ds = YP 2F10 + Tsinh"( 25+} +e RIM Then, {ta + 9 6=BEs he «ff 20-00) cm Pe =-4 0-4) 6-9, ef fh foila- MT ve) Then, {ie+ 5) fo +s de --} 56-204 « BY fa weak eB (ale FAO =~ (6-28 +8 + Bf alle -Tf Bae wt ateoh BASE Tl tad Bla Teal 4 *-(6-t safe ofa 74 2 | | a (3 i “(6 whale ‘il 4 “= (6-244 sft, Hi *5-) Sls) fg Bt; +4 deme (Ga gel — 5.608 1+f lee eereeehe Se Feosx+3sinx 'e nS ae Laue Pee oa Flog! = = Hig] 28 re = bog eaat te 4 sectx dx = dt dx= ‘Laas « SarSine= rosa ee taana, evaluate [4+ 5x10" fete Soak Sivan? 21an0 [esin20= 70 | “aad (iene) 25) tan eesceaes 2M 444 tan? +10tany Let us assume, tan f=! On differentiating both sides, we get, se $4 de=dt 2dt 2m sec? F = dr = 2d 2x 1+ tan? 2a : ee ae 22M 2x ae 4 tan? 5 +10 tan +4 ane _ uae) ee dt at at, eA =m) eB XS 7A + B= DAN ing the coefficients on both sides, waa ack IAtBaS= 7d) +a-s = Bes-$ sae} sexe $(r-20)4} Jase Sed ten 1 3/4 = 5 Wie =10 + $f. aa ds . peri +2) ea “[e-2{2}e - crew +2) ee [2 = VaR TER 1 + sayy 2)5 aa es PRLS Sees A we < py second gg 4 aKm #4 Fahl evowate J (yen) V3" : Lis, ‘sot, Given intesral a) Bei it “y ie tere ee eI = etd ve Scheer 5 +2 A 1 =f-t ee =- fis a — ‘ Aes 12 -1 watt, 2 UU Ge 2 Le, x+1=1 axel Differentiating on both sides, a= at de oe Tee Dae t3ee7 presse gan 3 const integral iss f SSE HEORESS de sine +3 cose +4 =A SL Q sinx +d cosy +5) +B @sinx +A cosx +5) +C AG cosx +4 (~sinx)) +B @ sinx + 4.cosx +5) +C =A (3 .cosx ~ 4 sinx) + B (3 sinx + 4 cosx +5) +C Dsinx +3 cost +4 = (4A + 3B)sinx + (3A + 4B) cosx + 5B+ C g the coefficients of sinx, cosx and constant terms on both sides, 3A+4B=3 (1) =4A+3B=2 ..Q) SB+C <4.) ‘equations (1) & (2), 3x(1) => 9A+12B = 9 4xQ) > -A6A+12B ituting A= Je inequation (1) stuting B= © in equation @) 5(33)+c=4 Zeon t4 jy = 1 f_Seosrtsinn 4.18 / 4 cosx +5 35) Ssinx + 4eosx +5 * 25. Ndr eee = se tog|3sins + oonn+ sleet 3 | same aeos +e te Fsmxt4oosxt5 > | 3 2 (1am? fig. piu ie 3 Guan} 4-4 tan +54 Stan ax pes a ee ‘ Sol. vate Ai ote ant BoD act roe sadeeeae | | oie 4 toa 3h ae ia Hate ze (m+ Sear “peer ape if 2 1 pelts et Pre +: iz: 12 2 Bog | 2438412 1-2 a5 es 2 ea = Slog [>in Foun f 1AQ30. Sol. Let 1,= [sin'x.x n= Noe aig n in ~); a 5 e fin Yysinx de = sin xa = sinYx(-cosx) - f (n—1) sin, sin? y.cosx + if (n=1) sin" x(1= = sin Ys.cosx (21 f sin == sins, cosx+(n—1) I, = > 1+ (0-1, =— sin*Ys.cosx+ (n > A.) == sala con + (gh > 1,= Sint x. cose a Forn=4 J = fointrar = “ J sin'x.dy = Hn cone positive Integer, n> 2 and deduce the valve of [ cot*xdx . J any26, March Le fesse = f cot -xcot*xdx = f cot"~*x(cosec?x - 1) dx = Jot" *xcosee%xax ~ f cot" ?xax aT a bef pop Gan= LOT 8 cot’x 3 EX + cotxtxte _—— ns _— % eee al oe ax, nbeing ap 1AQ33. ‘optain the reauction af det t ee aie AUS 1 Son. Given, 1,= f 084 | f f sin x(n — 1) 608" 2x fe eee woorcte | OE iy f cont 2 Ga "2x (1 — c0s*x) dx oe | = cost tx sink * a cost-izsins #072) fe Bee = coststxsint + (1-1) cost nde —(0-V) feo =I, OD : = J, = cost ixsins + (t | > pena = cosmzaine + (0-2 fa | Teatro , =e? When n= 4, _costtxsinx , GD y, cos'xsinx 5 3 3 aia _ cos'xsinx , 3 | cos’ is 4 - 4 Jeosts de = Deduction When n = 3 vate / esin(bx +¢)dx ; (a, b,¢ €R,b 4 0) onR. sin(bx+e)de=I say je know that f fl)g(x)dr= fd goddr f (dn f ous) ae jere fix) = &*, g(x) = sin(bx + 6) info odde= of sinter ae f [Ler] fsintoxoae| tc (bx +. ae “cos (bx + e)dr ~e*coalbx+e) af = = ee ee gino evoivote J x tie? ou Given integral sete pe solved using Triscan Asbrt7t = wehers~ age! ete aS S ERED * ory re"? > 12a crnerd) + BC) Andx=0 1 tae 2 472 Putx=-1 1=B(-1) (142 = 17 BCG) =Be-l Putx=-2 1=C(-2r-2+1) 1=2cac=4 Aah,B=-1,C=4 DEFINITE INTEGRALS / Joo (1+ tanx)atx ‘ Jog(l + tanadae mS os|( (& Safix) + (I= satiy fa (1+ tanxy og2 - log(1 + tanx)]dx a og2)dx — f log(1+ tanx)ax é - Cater 2 C aartome—! [- From equation (1)] x & = 2 Star = & (a aeres > I 4) Sinx+cosx Let und «1 ee Sinx + 608%" 2M ‘sinx + {ieee nx + cosx f 9+ T6sinac® = (-sinx))dx = at (cos x + sin x) d= dt and (Gin x - cos x)= sintx + cos*x ~ 2sinx cosx = #? 1-2sinx cosx =F 1-sindx=F : oe ref reas | @ ei -{ ther“! * eu | rHar* : sue «| hax eft = 1-3) ira Grae (1=sinx) sina ~ 5) (rramelioeme re ale a Jo fae ‘ fers 4a [From equation ( — = acos’9 + bsin®® d x= A (acos?0 + bsin'o) IX _ 9(cos0(-sind)) + 2bsindcos0 = -2sindcos0a + 2bsindcosd = -asin20 + bsin20 = a(1- sin) + bsin’0 -a = d-asin’O + bsin’0 - 4 = @-asin’® b- x = b- (acos*® + bsin’®) = b— acos*9 — b(1 - cos*®) = ba cos* ~ b + beos* 8 = (b= acos? iM b-x=(b-a)cos® b-2=(b-a)cos® 1=cos*8 e=0 simplified Method Put x= acos'0 + bsin’0 dd = (b= a)sin20 a0 Now (t= a) = acoso + bain’ — a = (b~a)sin’o b-x=b-acos’0 ~ bsin’0 = (b- Uli kok = Ome" = 0-0 Lite ee 3M 4 Given integral is, / ssin)t oy t= J cot er) n= x)sin 2) ae Treost=3)' ee co Tr eos(—%) Were Trost Sea feos) e080] pita T+ cos’) f_nsin’x xsin’y t= fe | abe 2 fe dx ~ I [Fromequation (1)] mam ses T+ costs * f (1=cos*x) sinx Le Troe aim Let, cosr = Differntiating on both sides, ~sinxdx = dt = sinxdx = - dt Upper Limit 1 cost Lower Limit: x =0 #= cos0 =1 1-4f[bs8fom = ae.) 7 = Btq;'—2ftan4 oh" mm = Fl-1-1-2ftan-' 1) - tan“'()]] “3b 2-2[#-4]]- [9-2-9] : I= Ft-2+R] fetta ge 2 en x= 0=9 tan0=0=>0=0 _ Tatog2 108-1 it {+ From equation (1)] yelstand=150-% 21= 1082 [%-9 1- 2H) fsa aee ! = f lop tan) «36 40 oefs = fern ete aces alee | é I= 5 f sinc costed é = Let, fle) = sin’x cos fe) = site -9) cos’ (n- 4) * sin?x costs =f) = fy) =fe-2) This is an even function. sin’xcos’xdx => [= poo) a This is in the form of, on [ sensentede = ESD 6 Lo aman (Second Yee") + (etpre)s nab. Ao deduce & ; 2 nat the area ofthe region bounded by ~ ea of the circle x! + y?= a", * uation of the ellipse is, a -a is symmetrical about X and Y axis as shown in the figure 2 - font sray-o] {e = @alze2) ae Ta? b cD = ellipse= 4(Area of OCD) = 4( fab) = nab om ing b =a in equation () ellipse Becomes iis 2 e 3] =2 t6vis 16) 1s} ven y= Diva and ¥ 206 - xis 64,/ V7 5 gus 7 a HIFFERERFTIAL eau : aaa agian (44 x5 «yaa (0 UE RH EER Gq URKGR a fate x, cl 4 eye tunte riba te thyeg Walenta (4) ariel (Hy, parte qe ars Artegeattvgs tee (Ly 6! wo a rod wotution for the giver differential equation. ve (antee De Oe oS jo tie teu ito then, 1 «aca +0+0 |: gop o | wenstiten ayes ina Domne diferent sepeosnare RE ene cisingree? ms 38h pa : = - | pe athen, ndtennangen tthe with pests | sen COUT , am | Sutetining the commending wees in| es A=, Dao Cad ~ | = loge log(l +2) ~log(l ~») = ogee | substituting the correspond } equation @). te See S ees u log-log 29 = loge ~ Vis the requis oa (+ y) x= Gam'y—ay. | LA@a?. Solve! Siar ; Yaya | AQMP: Solve me aero equation differential equation is, Ceo harnd oe hand ix (any ~ x\dy Sol. Given differential equation is, | O-29 fe + 21y = xia Dividing by (1-22) on both sides, 22, ae -$e-G 0 aim \ \ This is in a sin the form of + px= 0 Equation (1) is of the form, Bary-a 2) Comparing equations () and @), par = er om ‘Integrating factor (LF) =e!" aStenelie® apr) Tint 2h oy" 2A = aM “The general solution is, yan =foanarre 2 valerie > Sar aM Let(l-)=t s m equation (1) can be written as, > -2dxedt > tax--F . ‘Then, xan fyedite -e¢-246 = bef ches ze frna 1 ore yee 1x 1-x) e ye et) yo Visca istered ea the given differential equation: —=— “ $e ares Saeed m0) £- saris es Lert ayee - vane —_Pilerraing om Bet aden wth respect 12 ow Vdd ae ®t hE) Siutattuting the corresponding values in equation (1), Maas W204 me Wey » WL ROH ADE aT We detS veg ae WET yards ing on both sides, 243 Hews fa ® Consider, AY Ao) wrest at | gage Lyf 209y09 a aa a Partial fractions improper 4 1 Am Sy ave) Af ze Vanes 3* fGkaav A From equation (), 1 efi lh vdafoofa oa 1ferrrig {aA gmx be vd I+ P4404) m x46 W + log (40 + One {+29 4 tog (4 44.8545 8 (Ue 24) +5) w gy tk (op, Gane 08 (x +8y +5) w gy Wy From equation(2y) ; 8y4 10g (x + By 45) 0 NT TOR Ae + By 45) 0p By pind the song Athe chord formed by 24 f= Fonthe tine ncaa 9 cjnen equation of the cinche ie 22 y= at ie OF -# pose coms c™ 0/9) radius 1 8 Giwen equation of the Sine is xcos + ysines = 7 scones + pina 9 =O = paper a RT ao _ last eel joos60)+ sina0) — p\ ‘a+ sin “Re g Now d= fee sence cot 247 = TF a7? saa e lenin ct toe croraiercepaeymne cece eX" SY nara 22=0onthe ling, Thing 22 May 22 March yarn ven equation of circle is, gam x4 BW-2=0 Line, y= 4-3 “phe center of the above circle is, is the perpendicular distance from | 7» ye eee tore d= lave (ieee SEES 2 : Jaypee 1 Length of the chord = afe-# ae echord= 4.76 units: i de Length of th + y +190, cular alstance from the center £0 thelinex+ y “dis the perpen’ aye 1+) ys Se emetyital axteil, 6 y ee 2 Y = , d= 3y2 ing Length ofthe chord = 2/7 = 4 =2V(3¥- Gy?) eg =2y25 Length of the chord= 27 units, SAGA, Find he pole of 3x + 4y = 45 = Owith respect fo the x" + y*— 6x —By + 5 =0, $ol, Given equation of circle is, Ho3 _ =34- 445 at typ- 6-8 +5=0 «= @) s =e Here, 2¢ =~6, 2/8, ¢=5 45x, + 135 = -9x, - 12y, +15 reraraed 36x, - 12y, -120=0 Given line is 3x + 4y ~ 45 = 0 = Q 3x,-y,-10=0 Let P(x, y,) be the pole. @-3x@ => Equation of the polar of P wrt (1) is 4x, -39,=0 0) 9-28-30 =0 , +Yy, +80 +H) +/fy ty) +020 MAM t-3 (+x) dy ty) +5=0 (5-3) +, -4) ~3r, ~dy, +5=0 @) But the equations ‘he same straight line ©) & @) respectively Substitute ‘x,’ value in equation (4) 46) -3y, =0 14 the cree of the Ae nine nae o win tne vosraeny met BY te mona (2,4) Yo he eee ery equation of clvele toy wont at yl = 2d = dy +25 % 0) sent food cos Point, Pixy y¥)) = B,= 4) Ay 16-309 Br = Ay = 25-0 Sy= dy = 25" 0 By Ay = 25 yamx+ candihe cliclex’+y?= a? Intersectal: 28 (1 +m?) (a? = 2). nter (C) = (0,0), Radius (2) = 4 ngth of the chord(AB) = 2¥7 ~ pth of an fexey Soe re of tor Gens , wey <4 ‘Point, F 14 = dare conjugate lines with respect to th ~— . sand 2x+ky-8 x Ifthe lines £*¥ =>) other. Soa = - ast ms") smn) gst mf G aye ra 2 m=kn=-8 Here = 1m, 21m 7-5 b= 2m Stope comer (Q= Ce A= AD , pains () = Vea = IFIED Here r= 3 ‘The sng the corresponding values in equation (1), ye (731) (2)+108)] = (1) 1) + -)- C5) 21) +1) - 8) yt = 32+) =(-1-1+5) (-2-k+8) = = 6+3k=@)(6-K) 6 +3k= 18-3k= 6k=12 i Ee. 7 $AQ8, Find the equation ofthe fangent at the point 30° (Parametric valve of 6) e xt y+ 4x4 by ~39 = 0, ; Sol, Given equation of circle is, @ Xi y2+ 4x + 6y-39=0 Centre, C=(-g,-f) = (-2,-3). Li Radius, r= /g?+f¥—c = J(2)"+(3y2439 = (5p =2V/3 The equation of the tangent ata point ‘6’ Of the circle 2° + y+ 2gx + 2fy +em0iegs x (+g) cos +(y+/) sing =r E 2. (+2) 0s 30° + (y +3) sin 30° =2 15 = 613.2 y+, (Laos > ONS +449) 45 > Vix+y+342/3 45 =o 3 is? 220/5 2/83) Oe te rernecdese of tangent. ae #. Fincl the equation of the 9f langent parallel to i, Given equation of circte is, Wty! 2v+4y=0 Point. "0. ~1) The equation of tangent at (1) is, miuen! 12 K+ Vt 244 4% OF (3,1), Alas fits wre ecqlonon Mert ~() AO + VD~ FOr + Sy 00 mS W-y-x-342y-2"0 De 2x+y-5-0 Slope of the tangent is, "Here, ¢ = - 1; f= 2; radius, r= Vir4n0=/3 The equations of tangents to equation (1) are, yt fa m(x+ gtr vitme D yt2-- 26-14 Sire => yt2=-2Ax-1) +5 (V5) => 2tyt5=0 The equation of tangent at (3, ~ 1) is 2x +y-5=0 ‘The equation of tangent parallel 2x + y-5=0is 2x +y +5=0 Q10. Ifa point P Is moving such that the lengths of tangents drawn from ‘P' to the circles x? ix ~ by ~ 12 = 0 and x? + y? + 6x + 18y +26 = 0 are in the ratio 2: 3 then find the equation of the locus of P. March-t7 Given equations of circles are, S= x+y? 4x-6y-12=0 ~() Si= xi + y2 + 6x + 18y +26=0 ; 2 of tangent from P(x, y,) to $= 0s, gth of tangent from P(x, y,) to S’=0is,_ = Vai yx, #18y, +26 The contre lies on the X-axis = -fe0e fed stig ‘The required equation of circle is, ont) patos hough the point 2 ne SR ee sees > = 4g te=-13 & Iso ss through the point (4, co Martepscise ier razen ti > atce-d1 : Subemcng ‘equation (2) from equation (3), (Bg +0) - (-4¢ +0) = - 41 - (- 13) 67 = en = on et The equation of the required circle is eys2(s e+ 20) y- = _ 32+) 14r-67=0, SAQI2. Find the angle between tangents drawn from (3, 2) to the circle x2 + y2agx 4 Sol. 2 4b * — Given equation of the circle is, PtP 6x +4y-2=0, Point (P) = (x,, y,) = @, 2) Figure illustrates the with tangents drawn from point P. C=(3,-2) Radius ()= /g4f2—¢ = (3). (2a oye VOF4ED r= Vis oS oi ATE Fea aE = 19+4—- 1848-7 « ¥5,=1 Let’6’ be the angle between two tangents, ie 9_ vis Yaa at then fan Nic 8 I~tan?2 cos 9 = 12> V+ tan2f room ( =I5 _ ~14 7 THIS Tig; = SAQ14. Find the radical centre of the following circle. xt ¢ yt axn by #520, Xt y!=2x=dy-1 80, xt y= ox -2y=0. May. 24, Feb Aarch-24, Avg 2, May 1® Given system of circles are, Saxty4r-6y+5=0 S'extty'-2x-4y-1=0 S’ext+y-6x-2y=0 ‘The radical axis of the circles S=Oand S'= Ois, S-S'=0 > CP+Y— dr-6y45)— (2+ e-2x-4y-1)=0 Sol. => -2r-2y+6=0 = x+y-3=0 =) = nt Tadical axis ofthe circles §‘ =Oand "= Ois S-s’=0 = (P+ Y- 2x 4y-1)- (2+ -6x-2y) =0 = 4r-2y-1=0 (2) Solving equations (1) and (2) 4x Equation (1) > 4x + 4y-12=0 Equation (2) => 4x-2y-1=0 ++ a TT- ae Substituting y value in equation (1), xt 6 -3=0 2 Radical centre (x,y) =(2, 11) Suxtty'+3r+5yeq,, Saxty+x+3yeqy, ‘The common chord of two circles is their radical axis Equation of common chord is s-S'=0 4 = (A+ oF Far tSyt A) ek = x-y=0 Length of common chord = 2x For S=0, o-(S= one Ok The length of perpendicularfrom C= (GA) chords “ [a2 te Vere 2-3) (+ Cp? # Substituting the. ‘corresponding’ equation (3), wee! L=2 gat =2/4 =2x2-4 +: Length of common chord = 4. the equation and length of Chord” of the following s ey'eaxeayerno Ms yitaxt3y e200 system of circles are, eat ty + Det 2yt1=0 Saxtytdr+3y+2=0 Marensar as -S'=0 Pty 42r+2y+1)- (e+ +4e+3y+2)=0 $AQ17, Find the equationot he circle passing Tough the points of Inerection of x+y! Bx~ by 42180 MEF YE=2K=15 = 00nd (1,2). apt Sol. Given equations of circles are } + y-Be-6y+ 21-0 4 y'-2r-15=0 ‘The equation of circle passing through the Points of intersection of given circles is, Sty 88-220 h aay ‘As the equation (1) passes through the point (1,2), (12+ 2-8(1)-6(2) +21) + 1? +2°=2(1) -15)=0_ sant ne Substituting the value of in equation (1), = (tty -8r-6y+2e db ery-2 iz => 2? + 2y?- 16x - Ly + 424+ are coefficients of x,yandcis | — ion of chord 4, {@ep+Men+t eH seal “F w-[%} I lz | PRL Same ‘elrele which passes through the points (2,0), (0,2), SAO etiele nee ays Bes by eam 8 sol. Given equation of the circle Is, De avi a Se gy dO 4 se ¢ 2 2 Peewee w > vey =o a, The required circle passes through the points (2.0), (0, 2) The fandard equation of the cirele is, P4Y + ert BWyteno ) 20, qastuation @ passes through the point CP + OF +292) +270) +e=0 = 4gte+a-0 --B) (03, qeaustion @) passes through the point OF + QP +20) +270) +e=0 > Yrer4e0 ~@) Solving equations @) and (4), 4g+e+4=0 4ftc+4=0 S5F50) sae 6) Applying orthogonality proy equations (1) and (2), secretin (6) Substituting equation (5) |, 5g -6g-2e-4=9 =g-26-4=0 > gn-2-4 Substituting equation (7) in AC 2c-4) +4459 > > -Bc-16+c+4=0 ee => =%-12=0 Apr > %n-12 s = Th Substituting c= —!2 5 ¢™ 7 Anecaa Substituting the correspond: equation (2), The required equation of the circa) 70? + y}) ~ 8x By 19 =9, find the equation of the circ! ‘and having the Be wile en equation of the pty -4x+2y-7= Trere (6: f)~ @ Msc =-7 eee ty t2ext2ytc=0 Substituting the values of g, fir equation @) Bey + 22x + 2ABy ten 0 ; orthogonality property bet BS 2(-2X-D+0-3) = 0-7 B24 -3)-c-7 => 2=c-730"9 The required equation of the circle 20 { below, orthogonally xiey'-4x+ by #1050, xtty'+12y+6=0, “Given system of circle: sot - 4x + 6y +10 we4 yp +12y+6=0 ‘The standard equation of the circ! . ety t2gxt2fyrc=0 IM w= (I) @) Gince the circle passes through origin (0,0), 5 (OP + (0) + 29(0) +370) += 0 “en0 “Then, the equation is, Dart p+ 2x + 2fy=0 (3) 2-0, [Applying orthogonality property. be fations (1) and (3), Deg +f’) 6 2ig(-2) + /)] = 9+ 10 “22g + 3/) = 10 ) -6f+ 1 ty between orthogonality prope! cuts Mathemvaticnt() orthogonally the clicle xt + y!= ax + 2y=70 From Equation @)] (Second Yea) ‘Mar, aya Mar «) @ ) sn equuations (I) and (, = 1466 -> 17f-6=0 wim => 13/86 ny Substituting /= 4 inequation (t), = 4-0(4}+10=0 =p 4g-3410=0 = Age-7 = sce mim Subatituing gfe * > imesation sea yea dex t 2fy 20 \xva(h}yeo yogae yd ee Axrtyyereraye0 wim ‘The required equation of the circle aca r y= 7x 2y 20

You might also like