Let C_1 and C_2 be two circles touching each | vedclass

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(A) Let $r_1$ and $r_2$ be the radii of circles $C_1$ and $C_2$ respectively. Since $AB$ is the diameter of $C_1$,$AB = 2r_1$. Since $C_1$ and $C_2$ t

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$\frac{\sqrt{30}}{5}, \frac{3 \sqrt{30}}{10}$

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(A) Let $r_1$ and $r_2$ be the radii of circles $C_1$ and $C_2$ respectively. Since $AB$ is the diameter of $C_1$,$AB = 2r_1$. Since $C_1$ and $C_2$ touch externally at $A$,the line segment $BC$ passing through $A$ is the diameter of $C_2$,so $AC = 2r_2$. By the Power of a Point theorem for point $B$ with respect to circle $C_2$: $BA_2 	imes BA_3 = BA 	imes BC = (2r_1) 	imes (2r_1 + 2r_2) = 4r_1(r_1 + r_2)$. Given $BA_2 = 3$ and $BA_3 = 4$,we have $3 	imes 4 = 4r_1(r_1 + r_2)$,which simplifies to $r_1^2 + r_1r_2 = 3$ (Equation $i$). Let $M$ be the midpoint of chord $BA_1$ in $C_1$,so $BM = \frac{1}{2}BA_1 = 1$. Let $N$ be the midpoint of chord $A_2A_3$ in $C_2$,so $BN = BA_2 + \frac{1}{2}A_2A_3 = 3 + \frac{1}{2}(4-3) = 3.5 = \frac{7}{2}$. Considering the perpendiculars from centers $P$ and $Q$ to the secant line,we have similar triangles $	riangle BMP \sim 	riangle BNQ$. Thus,$\frac{BM}{BN} = \frac{BP}{BQ} = \frac{r_1}{2r_1 + r_2}$. Substituting values: $\frac{1}{7/2} = \frac{r_1}{2r_1 + r_2} \Rightarrow \frac{2}{7} = \frac{r_1}{2r_1 + r_2} \Rightarrow 4r_1 + 2r_2 = 7r_1 \Rightarrow 2r_2 = 3r_1$ (Equation $ii$). Substituting $r_2 = 1.5r_1$ into Equation $i$: $r_1^2 + r_1(1.5r_1) = 3 \Rightarrow 2.5r_1^2 = 3 \Rightarrow r_1^2 = \frac{3}{2.5} = \frac{6}{5} \Rightarrow r_1 = \sqrt{\frac{6}{5}} = \frac{\sqrt{30}}{5}$. Then $r_2 = 1.5 	imes \frac{\sqrt{30}}{5} = \frac{3}{2} 	imes \frac{\sqrt{30}}{5} = \frac{3\sqrt{30}}{10}$. Thus,the radii are $\frac{\sqrt{30}}{5}$ and $\frac{3\sqrt{30}}{10}$.

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