If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

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

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

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

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2 a x + 2 b y + (a^{2} + b^{2} + 4) = 0

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Solution

The correct option is D $2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0$
$x^{2} + y^{2} = 4 C e n t r e (0, 0) r a d i u s (r_{1}) = 2 L e t c o o r d i n a t e s b e (h, k) a s t h e c i r c l e s a r e o r t h o g o n a l . ∴ {r_{1}}^{2} + {r_{2}}^{2} = d^{2}$
$2^{2} + ({h - a)}^{2} + ({k - b)}^{2} = ({h - 0)}^{2} + ({k - 0)}^{2} 4 + h^{2} - 2 a h + a^{2} + k^{2} - 2 k b + b^{2} = h^{2} + k^{2} 2 a h + 2 k b - (a^{2} {+ b}^{2} + 4) = 0 ∴ 2 a x + 2 b y - (a^{2} {+ b}^{2} + 4) = 0 L o c u s o f t h e c e n t r e .$

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Similar questions

(a, b)

x^{2} + y^{2} = k^{2}

2 a x + 2 b y - (a^{2} + b^{2} + k^{2}) = 0

x^{2} + y^{2} + 2 a x - 2 b y + a^{2} - b^{2} = 0

The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

x^{2} + y^{2} + 2 a x + c^{2} = 0

x^{2} + y^{2} + 2 b y + c^{2} = 0

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

x^{2} + y^{2} + 2 a x - 2 b y + b^{2} = 0

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