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
- A \(6\)
- B \(7\)
- C \(4\)
- D \(5\)
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.…
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