AP EAMCET202119 Aug 2021Morning ShiftMathematicsCircleActual
Find the equation of the circle passing through ( 1 , - 2 ) and touching the x - axis at ( 3 , 0 ) .
Options
- Ax 2 + y 2 + 6 x − 4 y − 9 = 0
- Bx 2 + y 2 − 6 x − 4 y + 9 = 0
- Cx 2 + y 2 − 6 x − 4 y − 9 = 0
- Dx 2 + y 2 − 6 x + 4 y + 9 = 0
Correct answer
D. x 2 + y 2 − 6 x + 4 y + 9 = 0
Step-by-step solution
Let required circle equation is x - h 2 + y - k 2 = r 2 Given that circle touching the x   axis at 3 ,   0 . So x axis is a tangent to the given circle i.e perpendicular distance from centre to tangent must be equal to radius. ⇒ k = r x - h 2 + y - k 2 = k 2 It is passing through 3 ,   0   &   1 ,   -   2 ⇒ 9 + h 2 - 6 h   = 0 ⇒ h - 3 2 = 0   ⇒   h = 3 1 - 3 2 + - 2 - k 2 = k 2 ⇒ 4 + 4 + k 2 + 4 k = k 2 ⇒ 4 k + 8 = 0 ⇒ k