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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

  1. Ax - 9 2 + y + 8 2 = 25
  2. Bx - 9 2 + y - 8 2 = 25
  3. Cx + 8 2 + y + 8 2 = 25
  4. 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

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