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Maths-

General

Easy

Question

If a circle passes through the point (a, b) and cuts the circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals p to the power of 2 end exponent$ orthogonally, then the equation of the locus of its centre is‐

$x^{2} + y^{2} - 3 a x - 4 b y + (a^{2} + b^{2} - p^{2}) = 0$
$2 a x + 2 b y - (a^{2} - b^{2} + p^{2}) = 0$
$x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - p^{2}) = 0$
$2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$

The correct answer is: $2 a x plus 2 b y minus left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus p to the power of 2 end exponent right parenthesis equals 0$

To find locus for a given condition
Let the circle be $x 2 + y 2 + 2 g x + 2 f y + c = 0$ .
ince it cuts $x 2 + y 2 - k 2 = 0$ orthogonally therefore
$2 g (0) + 2 f (0) = c - k 2 = 0 ∴ c = k$ 2
Again the circle $C_{2}$ passes through $(a, b)$
$∴ a^{2} + b^{2} + 2 g a + 2 f b + k 2 = 0$
$∵ c = k$ 2
$∴$ Locus of centre $(- g, - f)$ is
$a 2 + b 2 + 2 a (- x) + 2 b (- y) + k 2 = 0$
or $2 a x + 2 b y - (a 2 + b 2 + k 2) = 0$ .

Hence locus of the given condition is optionD

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