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AP EAMCET · Maths · Circle

The least distance of the point \((10,7)\) from the circle \(x^2+y^2-4 x-2 y-20=0\) is

  1. A \(6\)
  2. B \(7\)
  3. C \(4\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(5\)

Step-by-step Solution

Detailed explanation

Given equation \(\left(x^2+y^2-4 x-2 y-20=0\right)\) So, \(C=(2,1)\) and radius \(=\sqrt{(g)^2+(f)^2-c}\) \(=\sqrt{(2)^2+(1)^2+20}\) Radius \(=5\) When substituted \(x=10\) and \(y=7\) in equation, then value becomes \((10)^2+(7)^2-4(10)-2(7)-20=75\) which is greater than zero.…
From AP EAMCET
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