If a circle passes through the point $(1, 2)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally then equation of the locus of its centre is the straight line $2 x + 4 y + 9 = 0$ .

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Solution

By $S_{1} + λ S_{2}$ the required circle is
$(x^{2} + y^{2} - 2 x + 6 y 6) + λ (x^{2} + y^{2} + 2 x - 6 y + 6) = 0$
or $(x^{2} + y^{2} - 6 y + 6) (1 + λ) + 2 x (λ - 1) = 0$
or $x^{2} + y^{2} + 2 x \frac{λ - 1}{λ + 1} - 6 y + 6 = 0 . . . . . (1)$
It cuts $x^{2} + y^{2} + 4 x + 6 y + 4 = 0$ orthogonally.
Hence $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
i.e. $2 \frac{λ - 1}{λ + 1} . 2 - 2.3 .3 = 6 + 4$
or $4 (λ - 1) = 28 (λ + 1)$
$(λ - 1 = 7 λ + 7 ∴ λ = - 4 / 3$
Hence the required circle by $(1)$ is
$x^{2} + y^{2} + 14 x - 6 y + 6 = 0$ .

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If a circle passes through the point (1, 2) and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the equation of the locus of its centre is

(a . b)

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

(a, b)

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

(3, 4)

x^{2} + y^{2} = 9

3 x + 4 y = λ

λ =

(a, b)

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

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