Let a circle passing through $(4, 0)$ touches the circle $x^{2} + y^{2} + 4 x - 6 y - 12 = 0$ , then radius of circle $C$ is:

5 \sqrt{2}

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\sqrt{57}

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5

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4

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Solution

The correct option is C $5$
Given,

$x^{2} + y^{2} + 4 x - 6 y - 12 = 0$

Equation of tangent at $(1, - 1)$

$x - y + 2 (x + 1) - 3 (y - 1) - 12 = 0$

$3 x - 4 y - 7 = 0$

Therefore the equation of circle is

$(x^{2} + y^{2} + 4 x - 6 y - 12) + λ (3 x - 4 y - 7) = 0$

It passes through $(4, 0)$ :

$(16 + 16 - 12) + λ (12 - 7) = 0$

$\Rightarrow 20 + λ (5) = 0$

$∴ λ = - 4$

$∴ (x^{2} + y^{2} + 4 x - 6 y - 12) - 4 (3 x - 4 y - 7) = 0$

or $x^{2} + y^{2} - 8 x + 10 y + 16 = 0$

Radius $= \sqrt{16 + 25 - 16} = 5$

Suggest Corrections

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