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
- A 5,10
- B 4,15
- C 5,15
- D 2,10
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 \(r=5\)
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