Let A(1, 2) be the centre and 3 be the radius | vedclass

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(A) Let $C_1 = A(1, 2)$ and $r_1 = 3$. Let $C_2 = B(-1, -1)$ and $r_2 = r$. The distance between the centres $d$ is given by $d^2 = (1 - (-1))^2 + (2

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(A) Let $C_1 = A(1, 2)$ and $r_1 = 3$. Let $C_2 = B(-1, -1)$ and $r_2 = r$. The distance between the centres $d$ is given by $d^2 = (1 - (-1))^2 + (2 - (-1))^2 = 2^2 + 3^2 = 4 + 9 = 13$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2}$. Given $	heta = \frac{\pi}{3}$,we have $\cos \frac{\pi}{3} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \frac{13 - 3^2 - r^2}{2 	imes 3 	imes r}$. $\frac{1}{2} = \frac{13 - 9 - r^2}{6r} = \frac{4 - r^2}{6r}$. $3r = 4 - r^2 \Rightarrow r^2 + 3r - 4 = 0$. $(r + 4)(r - 1) = 0$. Since $r$ is a radius,$r > 0$,so $r = 1$. Thus,there is only $1$ possible value for $r$.

Let $A(1, 2)$ be the centre and $3$ be the radius of a circle $S$. Let $B(-1, -1)$ be the centre and $r$ be the radius of another circle $S^{\prime}$. If $\frac{\pi}{3}$ is the angle between the circles $S$ and $S^{\prime}$,then the number of possible values of $r$ is

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