NTA Abhyas JEE Main2020MathematicsCirclePractice
The equation of the circle whose radius is 5 units and which touches the circle x 2 + y 2 - 2 x - 4 y - 20 = 0 externally at the point 5,5 is
Options
- Ax - 9 2 + y + 8 2 = 25
- Bx - 9 2 + y - 8 2 = 25
- Cx + 8 2 + y + 8 2 = 25
- Dx + 8 2 + y - 9 2 = 25
Correct answer
B. x - 9 2 + y - 8 2 = 25
Step-by-step solution
The equation of the given circle is x 2 + y 2 - 2 x - 4 y - 20 = 0 Its centre is C 1 1,2 and radius = 5 . The circle touches another circle of radius 5 externally at P 5,5 . Let its centre be C 2 α , β . Clearly, P 5,5 is the midpoint of C 1 C 2 . Therefore, α + 1 2 = 5 and β + 2 2 = 5 ⇒ α = 9 , β = 8 . Hence, the equation of the required circle is x - α 2 + y - β 2 = 5 2 or, x - 9 2 + y - 8 2 = 5 2