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MHT CET · Maths · Circle

The least distance of the point \(A(10,7)\) from the circle \(x^2+\mathrm{y}^2-4 x-2 \mathrm{y}-20=0\) is length of seg. AM . If \(\mathrm{MM}^{\prime}\) is the diameter of the circle, then the lengths of AM and \(\mathrm{AM}^{\prime}\) are respectively _______ , ______ units

  1. A 5,10
  2. B 4,15
  3. C 5,15
  4. D 2,10
Verified Solution

Answer & Solution

Correct Answer

(C) 5,15

Step-by-step Solution

Detailed explanation

Center of circle: \(x^2+y^2-4x-2y-20=0 \Rightarrow (x-2)^2+(y-1)^2 = 25\) Center \(C=(2,1)\), Radius \(r=5\)