Question

Let two circles $C_1$ and $C_2$ of radii 2 and 4 be tangent at point P and tangent to a common straight line (not passing through P) at points Q and R, then value of $PQ^2 + QR^2 + RP^2$ is-

• A. 48
• B. 56
• C. 64
• D. 72

Answer 1

Answer: (C)

64

Answer 2

Let $r_1$ and $r_2$ be the radii of the two circles $C_1$ and $C_2$, respectively. We are given $r_1 = 2$ and $r_2 = 4$. Let the common tangent line be L, and let Q and R be the points of tangency of $C_1$ and $C_2$ with L, respectively. The distance between the points of tangency Q and R on the common tangent line is given by the formula $QR = 2\sqrt{r_1r_2}$. So, $QR^2 = (2\sqrt{r_1r_2})^2 = 4r_1r_2$.

Let P be the point of tangency between the two circles. Let $O_1$ and $O_2$ be the centers of $C_1$ and $C_2$. The distance between the centers is $O_1O_2 = r_1 + r_2$. Consider a coordinate system where the line L is the x-axis. Let Q be at the origin $(0,0)$. Then $R$ is at $(QR, 0)$. The centers of the circles are $O_1 = (0, r_1)$ and $O_2 = (QR, r_2)$. The distance $O_1O_2^2 = (QR - 0)^2 + (r_2 - r_1)^2 = QR^2 + (r_2 - r_1)^2$. We also know $O_1O_2 = r_1 + r_2$, so $(r_1 + r_2)^2 = QR^2 + (r_2 - r_1)^2$. $r_1^2 + 2r_1r_2 + r_2^2 = QR^2 + r_2^2 - 2r_1r_2 + r_1^2$. $4r_1r_2 = QR^2$, which confirms $QR = 2\sqrt{r_1r_2}$.

Let P' be the projection of P onto the line L. The coordinates of P can be found using the section formula, as P divides the segment $O_1O_2$ in the ratio $r_1:r_2$. The y-coordinate of P (which is the height of P above L, let's call it $h$) is $h = \frac{r_2 y_1 + r_1 y_2}{r_1+r_2} = \frac{r_2 r_1 + r_1 r_2}{r_1+r_2} = \frac{2r_1r_2}{r_1+r_2}$. The x-coordinate of P' is $x_{P'} = \frac{r_2 x_1 + r_1 x_2}{r_1+r_2} = \frac{r_2 \cdot 0 + r_1 \cdot QR}{r_1+r_2} = \frac{r_1 QR}{r_1+r_2}$. This $x_{P'}$ is the distance $QP'$.

Now we can calculate $PQ^2$ using the Pythagorean theorem: $PQ^2 = (QP')^2 + h^2$. $PQ^2 = \left(\frac{r_1 QR}{r_1+r_2}\right)^2 + \left(\frac{2r_1r_2}{r_1+r_2}\right)^2$. Since $QR^2 = 4r_1r_2$, we have: $PQ^2 = \frac{r_1^2 (4r_1r_2)}{(r_1+r_2)^2} + \frac{4r_1^2r_2^2}{(r_1+r_2)^2} = \frac{4r_1^3r_2 + 4r_1^2r_2^2}{(r_1+r_2)^2}$.

Similarly, the distance $RP'$ is $QR - QP' = 2\sqrt{r_1r_2} - \frac{r_1 2\sqrt{r_1r_2}}{r_1+r_2} = 2\sqrt{r_1r_2} \left(1 - \frac{r_1}{r_1+r_2}\right) = 2\sqrt{r_1r_2} \left(\frac{r_1+r_2-r_1}{r_1+r_2}\right) = \frac{2r_2\sqrt{r_1r_2}}{r_1+r_2}$. So $RP' = \frac{r_2 QR}{r_1+r_2}$. Then $RP^2 = (RP')^2 + h^2$. $RP^2 = \left(\frac{r_2 QR}{r_1+r_2}\right)^2 + \left(\frac{2r_1r_2}{r_1+r_2}\right)^2 = \frac{r_2^2 (4r_1r_2)}{(r_1+r_2)^2} + \frac{4r_1^2r_2^2}{(r_1+r_2)^2} = \frac{4r_1r_2^3 + 4r_1^2r_2^2}{(r_1+r_2)^2}$.

Now, let's sum $PQ^2 + QR^2 + RP^2$: $PQ^2 + RP^2 = \frac{4r_1^3r_2 + 4r_1^2r_2^2 + 4r_1r_2^3 + 4r_1^2r_2^2}{(r_1+r_2)^2} = \frac{4r_1r_2(r_1^2 + 2r_1r_2 + r_2^2)}{(r_1+r_2)^2} = \frac{4r_1r_2(r_1+r_2)^2}{(r_1+r_2)^2} = 4r_1r_2$. So, $PQ^2 + QR^2 + RP^2 = (PQ^2 + RP^2) + QR^2 = 4r_1r_2 + 4r_1r_2 = 8r_1r_2$.

Given $r_1 = 2$ and $r_2 = 4$: $PQ^2 + QR^2 + RP^2 = 8 \times 2 \times 4 = 64$.