Let two circles C_1 and C_2 of radii r_1 = 2 | vedclass

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(C) Let the radii of the two circles be $r_1 = 2$ and $r_2 = 4$. The distance between the centers is $d = r_1 + r_2 = 2 + 4 = 6$.In the triangle $	ri

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$64$

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(C) Let the radii of the two circles be $r_1 = 2$ and $r_2 = 4$. The distance between the centers is $d = r_1 + r_2 = 2 + 4 = 6$.In the triangle $	riangle QPR$,the angle $\angle QPR = 90^\circ$ because the common tangent at $P$ bisects the angle $\angle QPR$.Thus,$QR^2 = PQ^2 + PR^2$.The length of the common external tangent $QR$ is given by $\sqrt{d^2 - (r_2 - r_1)^2} = \sqrt{6^2 - (4 - 2)^2} = \sqrt{36 - 4} = \sqrt{32}$.So,$QR^2 = 32$.We need to find $PQ^2 + QR^2 + RP^2 = (PQ^2 + PR^2) + QR^2 = QR^2 + QR^2 = 2 	imes QR^2$.Substituting the value,we get $2 	imes 32 = 64$.

Let two circles $C_1$ and $C_2$ of radii $r_1 = 2$ and $r_2 = 4$ be tangent to each other at point $P$ and tangent to a common straight line (not passing through $P$) at points $Q$ and $R$ respectively. Then the value of $PQ^2 + QR^2 + RP^2$ is:

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