Circles C_1 and C_2,of radii r and R respecti | vedclass

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(B) Let $O$ be the center of circle $C_1$ and $O'$ be the center of circle $C_2$. Let the line joining the centers be the $x$-axis. Since the circles

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$45^{\circ}$

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(B) Let $O$ be the center of circle $C_1$ and $O'$ be the center of circle $C_2$. Let the line joining the centers be the $x$-axis. Since the circles touch each other,the distance between their centers is $R-r$. Let $M$ be the projection of $P$ onto the line $OO'$. Since $l$ is tangent to $C_1$ at $P$,$PM \perp OO'$,so $PM = r$. Let $N$ be the projection of $B$ onto the line $OO'$. Since $l$ is parallel to $OO'$,$BN = PM = r$. In $	riangle O'NB$,$O'B = R$ and $BN = r$. Given $R^2 = 2r^2$,we have $R = \sqrt{2}r$. Then $O'N = \sqrt{O'B^2 - BN^2} = \sqrt{2r^2 - r^2} = r$. Since $O'N = BN = r$,$	riangle O'NB$ is an isosceles right-angled triangle,so $\angle BO'N = 45^{\circ}$. Similarly,$\angle AO'M = 45^{\circ}$. Thus,$\angle AO'B = 180^{\circ} - (45^{\circ} + 45^{\circ}) = 90^{\circ}$. The angle subtended by the chord $AB$ at the center $O'$ is $90^{\circ}$. The angle subtended by the same chord $AB$ at any point $O$ on the circumference of $C_2$ is half the angle subtended at the center. Therefore,$\angle AOB = \frac{1}{2} \angle AO'B = \frac{1}{2} 	imes 90^{\circ} = 45^{\circ}$.

Circles $C_1$ and $C_2$,of radii $r$ and $R$ respectively,touch each other as shown in the figure. The line $l$,which is parallel to the line joining the centres of $C_1$ and $C_2$,is tangent to $C_1$ at $P$ and intersects $C_2$ at $A$ and $B$. If $R^2=2r^2$,then $\angle AOB$ equals

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