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Number of Common Tangents to Two Circles in Different Conditions

The common ta...

Question

The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :

A

(- 6, 4)

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B

(6, - 2)

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C

(4, - 2)

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D

(- 4, 6)

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Solution

The correct option is B $(6, - 2)$
$x^{2} + y^{2} = 4$
Centre is $C_{1} (0, 0)$ and radius $r_{1} = 2$
$x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
Centre is $C_{2} (- 3, - 4)$ and radius $r_{2} = 7$

Distance between centres $C_{1} C_{2} = 5 = | r_{1} - r_{2} |$
Therefore, the circles touch internally.
Therefore, equation of common tangents is
$2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + (c_{1} - c_{2}) = 0$
$\Rightarrow 6 x + 4 y = 20$
or, $3 x + 2 y = 10$
$(6, - 2)$ lies on the above common tangent.

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