Let the point B be the reflection of the poin | vedclass

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(A) The distance of point $A(2,3)$ from the line $8x-6y-23=0$ is given by $d = \frac{|8(2)-6(3)-23|}{\sqrt{8^2+(-6)^2}} = \frac{|16-18-23|}{\sqrt{64+3

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$10$

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(A) The distance of point $A(2,3)$ from the line $8x-6y-23=0$ is given by $d = \frac{|8(2)-6(3)-23|}{\sqrt{8^2+(-6)^2}} = \frac{|16-18-23|}{\sqrt{64+36}} = \frac{|-25|}{10} = 2.5 = \frac{5}{2}$.Since $B$ is the reflection of $A$ in the line,the distance $AB = 2d = 2 	imes \frac{5}{2} = 5$.Let $r_A = 2$ and $r_B = 1$ be the radii of circles $\Gamma_A$ and $\Gamma_B$ respectively.Let $C$ be the point of intersection of the common external tangent $T$ and the line $AB$. By the property of similar triangles formed by the centers and the tangent point,we have $\frac{CA}{CB} = \frac{r_A}{r_B} = \frac{2}{1}$.This implies $CA = 2CB$.Since $C, B, A$ are collinear and $B$ lies between $A$ and $C$ (as $r_A > r_B$),we have $CA = CB + AB$.Substituting $CB = \frac{CA}{2}$,we get $CA = \frac{CA}{2} + AB$.$\Rightarrow \frac{CA}{2} = AB = 5$.$\Rightarrow CA = 10$.

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