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Radii, Chord, Diameter and Circumference of the Circle

If 6, - 3 i...

Question

If $(6, - 3)$ is the one extremity of diameter to the circle $x^{2} + y^{2} - 3 x + 8 y - 4 = 0$ then its other extremity is -

(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 B $(- 3, - 5)$
Centre of the circle is $(\frac{3}{2}, - 4)$
Other end of the diameter is
$\frac{x + 6}{2} = \frac{3}{2}$
$x = 3 - 6 = - 3$
$\frac{y - 3}{2} = - 4$
$y = - 8 + 3 = - 5$

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