AP EAMCET20224 Jul 2022Evening ShiftMathematicsCircleActual
A circle has its centre in the first quadrant and passes through 2 , 3 . If this circle makes intercepts of length 3 and 4 respectively on x = 2 and y = 3 , its equation is
Options
- Ax 2 + y 2 + 3 x - 5 y + 8 = 0
- Bx 2 + y 2 - 4 x - 6 y + 13 = 0
- Cx 2 + y 2 - 6 x - 8 y + 23 = 0
- Dx 2 + y 2 - 8 x - 9 y + 30 = 0
Correct answer
D. x 2 + y 2 - 8 x - 9 y + 30 = 0
Step-by-step solution
Consider a circle with centre be a , b . So the equation of circle will be, x 2 + y 2 - 2 a x - 2 b y + c = 0             ⋯ 1 In the question it is given that the circle passes through 2 , 3 . Therefore, substituting x = 2 and y = 3 respectively in the equation 1 we get, 2 2 + 3 2 - 2 a × 2 - 2 b × 3 + c = 0 ⇒ 13 - 4 a - 6 b + c = 0 ∴     c = 4 a + 6 b - 13 So now putting the value of c in the equation 1 we get, x 2 + y 2 - 2 a x - 2 b y + 4 a + 6