‘A conic section is the intersection of a plane and a cone. If the plane does not pass through the vertex of the cone, then the conic sections formed are a cirde, an ellipse, a parabola, and hyperbola (Domingo, Z@Z7F The standard equation of a circle can be derived 4 =J@= WPF OK Since, the distance from the center to point is radius (), then we can say that, rv =V@— WFO BE VOW FOF =r Squaring both sides of the equation leads to =? + 0-6)? ‘Therefore, the Standard fort OF Eh6 €GUatOA GFA EEE whose center is at (h, k) and whose radius is risgiven by (— 1)? + Cy — )* = 1*, where r> 0. reistne set or ai points ma piane tnatare equiaistanr trom a nxea point canea. the center. The distance from the center to any point on the circle is called the radius of the circle denoted by r, where r > 0 (Albay, 2016), ce G6 +3? 15 Com tot = “The given equation can be written as Sano a. 8 FO 9 04 etree ‘The given equation can he written as Katha aes ee DE +O -BD2= 8 Gam +o w? i ata DE +o THE GENERAL FORM OF THE EQUATION OFA CIRCLE) ‘The equation of a circle can be written in another form by expanding the binomials on the left side of the equation and then combining similar terms, Thus, Ce my? +0 —* xt = 2h + A? + y? = 2key AE rt XP yh Bhi Zh eI ‘The last equation can be written in che general form of the equation of a circie, that is, xt y'tDx+ by + F~0. 10N TRANSFORMING THE EQUAT OTHE EE GENERAL FORM T ‘The next examples present the steps in writing the equation of a circle from the general form to the standard form,Example 4; Determine the standard farm of the equation of the circle defined by RP bye Lax t dy +340. Solution: Group all the x and yterms and xt ae ty? 4 tay = 34 transpose the constant term to the Fight side of the equation. Complete the square of the x terms. Gt = 1a) +O? #149) = —34 (Recall: Completing the square) Complete the square of the y terms. 3444944 Adding 49 to both sides of the equation will keep its balance. Rewrite the perfect square trinomial to a square of binomial (on the lett side) and simplify the right side of the equation, ‘Therefore, the standard form of the equation of the circle is G27) G7 S64 G2 14x 449) + OF +a) = G7 +O +n? Example 5: Determine the standard form of the equation of the circle defined by x? +y? Ax + l0y—13 = 0. Solution: Group all the x and yterms and transpose the constant ‘term to the right side of the equation. Complete the square of the x terms. (x? — 4x) +O + Loy) = -13 (Recall: Completing the square) Complete the square of the y terms. 1344425 Adding 49 to both sides of the equation will keep its balance. Rewrite the perfect square trinomial to a square of binomial (on the left side) and simplify the rightside of the equation. Therefore, the standard form ofthe equation of the circle is (@= 27 #@#S)2 S16, GRAPH OF A CIRCLE ‘The rectangular coordinate system is used to sketch the graph ofa circle. The graph provides a clear view ofits center and radius. Consider the figure below: x? ty? — 4x4 10y-13 = G? = 4 + 4) + (y? + 10y +25) (= 2)? +O +5)' =16 Fig 1.1 A circle whose cont&ris at (2,6) andrExample 6: Sketch the graph of circle given the fo a +ytHO b GHD+O4+3%=5 © Genter at the origin and Solution: 2 The coordinates of the center and radiuis are required when graphing a circle. Given the standard form Ge = hy? + 19? = 17, = @ ‘Therefore, h=2, k= Oand r= 3. Thus, Note: Radius is the distance from center to any point on the circle Ifthecenter ofthe circle is atthe origin, then the coordinates ofthe conter is (0,0). Radius is 4 units In ength, this means thae the distance from the center to any point on the cele s 4 units. A What Have I Learned so Far? a ‘Sketch the graph of the circle given the following: 6 +52 +O +3)? G1? +GO-4)7=1 x+y? Center at (1,4) andr Center at the origin and r 2 4. Progress Check / Learning Activities / Exercises 1. Complete the table.I. Given the general form of equation, a) give the standard form of equation of a circle and (b) sketch 1. xt ty? 46x —4y412=0 2. xt +y#— 2x —12y +36 B.oXt 474 Lex Hay 15 Determine the equation that represents the graph ofa circleGiven the center and the radius, it would be easy to determine the equation of the circle. Example 1: Determine the standard equation of the circle given the coordinates of its center and the length of its radius. Center at @,—3) andr = 3 b. Centerat GD andr = 6 © Center atthe origin andr = V7 & Center (0.5) andr e. Center (—1,0) and r = 2V3_ Solution ‘The standard form of the equation of a circle is (x —h)? + G— 4)? =r. To get the standard equation of the circle, simply substitute the values of h, and rin the equation, then simplify. a. Sinceh = 2,k = —3,andr = 3,then the standard form of the equation of the circle is, Gas” andr = 6,then the standard form of the equation of the circle is, oo b. Since h Note that the center is at the origin, hence, the coordinates ofthe center is (0, 0). Therefore, we can say that h = 0,k = Oandr = V7. The standard form of the equation of the circle is @&-07 + ve Y = 7)? 4, then the standard form ofthe equation of the circle is (x= or + & - 2 ‘ e. Since h = —1,k = 0,andr = 2V3, then the standard form of the equation of the circle is CDF 0 5 (" d. Since h 5.andr Example 2: Given the standard form of the equation, find the coordinates of the center and the radius of each of the following circles. a (+9? + (y— 1D? =25 b. xt +y2=49 c @& +6) ty? =15 dx? + (y— 3)? = 64 © G+ +072" Solution: ‘The standard form of the equation ofa circle is (x —h)? + (y — 1)? = r2. To get the coordinates of the circle, consider the values that is being subtracted to x and y. To get the radius, simply get the square root of the constant on the right side of the equation, a @ +924 (9-2 =25 @-W?+Q-b2=7? ‘The given equation can be written as ss Gales ate ge bo x+y? =49 Wy? +Q- bP = 7? ‘The given equation can be written as Se o ie Ds =