If $(6, - 3)$ is the one extremity of diameter to the circle $x^{2} + y^{2} - 3 x + 8 y - 3 = 0$ then its other extremity is -

(\frac{3}{2}, - 4)

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(- 3, - 5)

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(3, - 5)

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(3, 5)

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Solution

The correct option is C $(- 3, - 5)$
Given $(6, - 3)$ is one end of diameter

Given circle equation is $x^{2} + y^{2} - 3 x + 8 y - 3 = 0$

we know that for circle equation $a x^{2} + b y^{2} + 2 h x y + 2 g x + 2 f y + c$ the center is $(- g, - f)$

$⟹$ center is $(\frac{3}{2}, - 4)$

Let the other end be $(x, y)$

Centre is mid point of both ends

$⟹ \frac{x + 6}{2} = \frac{3}{2}, \frac{y - 3}{2} = - 4 ⟹ x = - 3, y = - 5$

$∴$ the other end is $(- 4, - 5)$

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