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

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10-1.Circle and System of Circles

normal

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

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

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

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

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

Solution

Let $A, B$ be the centers of the two circles. Slope of the common tangent $=-\frac{12}{5}$ $\therefore$ Slope of $A B$ is $\tan \theta=-\frac{1}{-\frac{12}{5}}=\frac{5}{12}$

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)

$\therefore$ Coordinates of $A$ and $B$ are $(1 \pm 13 \cos \theta,-1 \pm 13 \sin \theta)$,

where $\tan \theta=\frac{5}{12}$

$\Rightarrow\left(1 \pm 13 \cdot \frac{12}{13},-1 \pm 13 \frac{5}{13}\right) \Rightarrow(1 \pm 12,-1 \pm 5)$

$\Rightarrow(13,4)$ or (-11,-6)

Standard 11

Mathematics

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Solution

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