If two spheres of radii r_1 and r_2 cut ortho | vedclass

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(D) Let the two spheres have centers $O$ and $O'$ and radii $r_1$ and $r_2$ respectively. Since the spheres cut orthogonally,the angle between the rad

Vedclass · Accepted Answer

$\frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}}$

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(D) Let the two spheres have centers $O$ and $O'$ and radii $r_1$ and $r_2$ respectively. Since the spheres cut orthogonally,the angle between the radii at the point of intersection $P$ is $90^\circ$. Let $C$ be the center of the common circle and $r$ be its radius. In $\Delta OPC$,the angle $\angle OPC = 	heta$. Thus,$\cos 	heta = \frac{r}{r_1}$. In $\Delta O'PC$,the angle $\angle O'PC = 90^\circ - 	heta$. Thus,$\sin 	heta = \cos(90^\circ - 	heta) = \frac{r}{r_2}$. Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have: $\left(\frac{r}{r_1}ight)^2 + \left(\frac{r}{r_2}ight)^2 = 1$ $r^2 \left(\frac{1}{r_1^2} + \frac{1}{r_2^2}ight) = 1$ $r^2 \left(\frac{r_1^2 + r_2^2}{r_1^2 r_2^2}ight) = 1$ $r^2 = \frac{r_1^2 r_2^2}{r_1^2 + r_2^2}$ $r = \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}}$

If two spheres of radii $r_1$ and $r_2$ cut orthogonally,then the radius of the common circle is

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