Two circles each of radius 5 text{ units} tou | vedclass

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(A) The common tangent is $4x + 3y = 10$. Its slope is $m = -\frac{4}{3}$.Since the line joining the centres is perpendicular to the tangent at the po

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$40$

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(A) The common tangent is $4x + 3y = 10$. Its slope is $m = -\frac{4}{3}$.Since the line joining the centres is perpendicular to the tangent at the point of contact $(1, 2)$,the slope of the line joining the centres is $m' = -\frac{1}{m} = \frac{3}{4}$.Let the angle this line makes with the $x$-axis be $	heta$,so $	an 	heta = \frac{3}{4}$.This implies $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The centres $C_{1}$ and $C_{2}$ are at a distance of $5 	ext{ units}$ from $(1, 2)$ along this line.Using the parametric form of a line,the coordinates of the centres are $(x, y) = (1 \pm 5 \cos 	heta, 2 \pm 5 \sin 	heta)$.$(x, y) = (1 \pm 5(\frac{4}{5}), 2 \pm 5(\frac{3}{5})) = (1 \pm 4, 2 \pm 3)$.Thus,the centres are $(1 + 4, 2 + 3) = (5, 5)$ and $(1 - 4, 2 - 3) = (-3, -1)$.So,$(\alpha, \beta) = (5, 5)$ and $(\gamma, \delta) = (-3, -1)$.Then,$|(\alpha + \beta)(\gamma + \delta)| = |(5 + 5)(-3 - 1)| = |(10)(-4)| = |-40| = 40$.

Two circles each of radius $5 \text{ units}$ touch each other at the point $(1, 2)$. If the equation of their common tangent is $4x + 3y = 10$,and $C_{1}(\alpha, \beta)$ and $C_{2}(\gamma, \delta)$,$C_{1} \neq C_{2}$ are their centres,then $|(\alpha + \beta)(\gamma + \delta)|$ is equal to .... .

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