The centre of the circle $x^{2} + y^{2} + 10 x - 20 y + 100 = 0$ is

(5, 10)

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(- 5, 10)

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(- 5, - 10)

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(5, - 10)

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Solution

The correct option is B $(- 5, 10)$
Given the equation of the circle is
$x^{2} + y^{2} + 10 x - 20 y + 100 = 0$
or, $x^{2} + 10 x + 25 + y^{2} - 20 y + 100 = 25$
or, $(x + 5)^{2} + (y - 10)^{2} = 5^{2}$
From this equation it is clear that the centre is $(- 5, 10)$ .

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Find the value of k if x + y + 5 = 0 is a tangent to the circle $x^{2} + y^{2} + 10 x + 2 k y + 10 = 0$

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