Question

There are two circles $C_1$ and $C_2$ of radii $3$ and $8$ units respectively. The common internal tangent $T$ touches the circles at points $P_1$ and $P_2$ respectively. The line joining the centers of the circles intersects $T$ at $X$. The distance of $X$ from the center of the smaller circle is $5$ units. What is the length of the line segment $P_1P_2$?

• A. $\le 13$
• B. $>13$ and $\le 14$
• C. $>14$ and $\le 15$
• D. $>15$ and $\le 16$
• E. $>16$

Hint:

For two circles with center distance $d$, the internal common tangent segment between tangency points has length $\sqrt{d^2-(r_1+r_2)^2}$. Use similarity from the homothety point on the line of centers to find $d$ when $O_1X$ or $O_2X$ is known.

Answer 1

The Correct Option is C

Step 1: Geometry and similar triangles.
Let the centers be $O_1$ (radius $r_1=3$) and $O_2$ (radius $r_2=8$).
$O_1P_1 \perp T$ and $O_2P_2 \perp T$, and $O_1O_2$ meets $T$ at $X$.
In right triangles $\triangle O_1XP_1$ and $\triangle O_2XP_2$, the acute angle at $X$ is common; hence the triangles are similar. Therefore, \[ \frac{r_1}{O_1X}=\frac{r_2}{O_2X}⇒ O_2X=\frac{r_2}{r_1}\,O_1X=\frac{8}{3} 5=\frac{40}{3}. \] Thus the distance between the centers is \[ d=O_1O_2=O_1X+O_2X=5+\frac{40}{3}=\frac{55}{3}. \] Step 2: Length of the internal common tangent $P_1P_2$.
For a common internal tangent, \[ P_1P_2=\sqrt{\,d^2-(r_1+r_2)^2\,}. \] With $d=\dfrac{55}{3}$ and $r_1+r_2=11$, \[ P_1P_2=\sqrt{\left(\frac{55}{3}\right)^2-11^2} =\sqrt{\frac{3025-1089}{9}} =\sqrt{\frac{1936}{9}} =\frac{44}{3}\approx 14.67. \] \[ \boxed{P_1P_2=\dfrac{44}{3}\ \text{units}\ (\approx 14.67)}⇒ \text{falls in }(14,15]. \]