If the equation of the common tangent at the | vedclass

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Question

Let $A, B$ be the centers of the two circles. Slope of the common tangent $=-\frac{12}{5}$ $	herefore$ Slope of $A B$ is $	an 	heta=-\frac{1}{-\fra

Vedclass · Accepted Answer

$(13, 4), (-11, -6)$

Vedclass · Answer

Let $A, B$ be the centers of the two circles. Slope of the common tangent $=-\frac{12}{5}$ $	herefore$ Slope of $A B$ is $	an 	heta=-\frac{1}{-\frac{12}{5}}=\frac{5}{12}$

The point (1,-1) lies on the line $A B$ and the points $A$ and $B$ are at a distance 13 from the point (1,-1)

$	herefore$ Coordinates of $A$ and $B$ are $(1 \pm 13 \cos 	heta,-1 \pm 13 \sin 	heta)$,

where $	an 	heta=\frac{5}{12}$

$\Rightarrow\left(1 \pm 13 \cdot \frac{12}{13},-1 \pm 13 \frac{5}{13}ight) \Rightarrow(1 \pm 12,-1 \pm 5)$

$\Rightarrow(13,4)$ or (-11,-6)

If the equation of the common tangent at the point $(1, -1)$ to the two circles, each of radius $13$, is $12x + 5y -7 = 0$, then the centre of the two circles are

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