AP EAMCET · Maths · Circle
The sum of the minimum and maximum distances of the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
- A 20
- B 16
- C 12
- D 64
Answer & Solution
Correct Answer
(A) 20
Step-by-step Solution
Detailed explanation
Circle center \((-2, 5)\). Radius \(r = \sqrt{2^2+(-5)^2-(-7)} = \sqrt{36} = 6\) Distance \(d\) from \((4,-3)\) to \((-2,5)\): \(d=\sqrt{(4-(-2))^2+(-3-5)^2}\) \(d=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10\) Sum of minimum and maximum distances: \(2d = 2 \times 10 = 20\)
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