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Tangent

Equation of t...

Question

Equation of the circle whose radius is 3 and which touches the circle $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ internally at the point $(- 1, - 1)$ , is

$5 x^{2} + 5 y^{2} - 8 x - 14 y - 32 = 0$

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$5 x^{2} + 5 y^{2} + 8 x - 14 y - 32 = 0$

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$5 x^{2} + 5 y^{2} + 8 x + 14 y - 32 = 0$

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$5 x^{2} + 5 y 2 - 8 x + 14 y - 32 = 0$

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Solution

The correct option is A
$5 x^{2} + 5 y^{2} - 8 x - 14 y - 32 = 0$

$C_{1} : (x_{1}, y_{1}), r_{1} = 3 C_{2} : (2, 3), r_{2} = 5$

Point P divides $C_{2} C_{1}$ externally in the ratio of their radius
$C_{2} P : C_{1} P = - 5 : 3$
or $C_{1} C_{2} : C_{1} P = 2 : 3$
co-ordinates of $C_{1} = (\frac{4}{5}, \frac{7}{5})$
Equation of circle is
${(x - \frac{4}{5})}^{2} + {(y - \frac{7}{5})}^{2} = 9 \Rightarrow 5 x^{2} + 5 y^{2} - 8 x - 14 y - 32 = 0$

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