Suppose that the equation of the circle having $(- 3, 5)$ and $(5, - 1)$ as end point of a diameter is $(x - a)^{2} + (y - b)^{2} = r^{2}$ . Then $a + b + r, (r > 0)$ is?

8

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Solution

The correct option is A $8$
$\begin{matrix} C = [(\frac{3 + 5}{2})], [(\frac{5 - 1}{2})] (3, 5) (5, - 1) C = [1, 2], [∵ i f e n d p o i n t s o f d i a m e t e r g i v e n], [T h e n i t s m i d p o i n t i s c e n t r e o f c i r c l e] T h e n i t s e q u a t i o n i s (i) (x - 1)^{2} + (y - 2)^{2} = (r)^{2} (i i) (x - a)^{2} + (y - b)^{2} = r^{2} C o m p e a r i n g t h e s e t w o e r_{s}^{h} a = 1, b = 2, r = 5 ∴ a + b + r = 1 + 2 + 5 = 8 \end{matrix}$

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A

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