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

8

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9

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10

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Solution

The correct option is A $8$
$(- 3, 5)$ and $(5, - 1)$ are end point of the diameter.
So;

$center = (\frac{- 3 + 5}{2}; \frac{5 - 1}{2}) (mid-point)$
$= (1, 2) = (a, b)$

Distance between $(- 3, 5)$ and $(5, - 1) =$ diameter of circle

$diameter = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$
$= \sqrt{(5 + 3)^{2} + (- 1 - 5)^{2}}$
$\Rightarrow \sqrt{64 + 36} = \sqrt{100} = 10$
So, radius $= r = \frac{10}{2} = 5$
Now; $a + b + r = 1 + 2 + 5 = 8$

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