+39 The two circles below are externally tangent. A common external tangent intersects line at Find line PQ at R. Find QR.
There are two circles touching, circle 1 has a point in the middle that extends to the border of the circle that's length is 12(P). Same with the other circle (Q)but its length is 8.
+179 Let $O_1$ and $O_2$ be the centers of circles $P$ and $Q$ respectively. Let $A$ and $B$ be the points where line $PQ$ intersects the circles $P$ and $Q$ respectively.
Let $C$ be the midpoint of $\overline{AB}$. Let $D$ be the foot of the perpendicular from $O_2$ to $\overline{PQ}$, and let $E$ be the foot of the perpendicular from $O_1$ to $\overline{O_2D}$.
Since $\overline{O_1E}$ and $\overline{O_2D}$ are both perpendicular to $\overline{PQ}$, $O_1EDO_2$ is a rectangle. Therefore, $O_1E=O_2D$.
Since $O_1O_2=12+8=20$, we know $O_1E=O_2D=10$. Because $\angle O_1AQ=\angle O_2BQ=90^\circ$, angles $O_1EA$ and $O_2DE$ are both right angles.
Thus, by the Pythagorean Theorem on $\triangle O_1AE$, \[AE=\sqrt{12^2-10^2}=2\sqrt{7}\]
and by the Pythagorean Theorem on $\triangle O_2DE$, \[ED=\sqrt{20^2-10^2}=10\sqrt3.\]
Finally, we find the length of $\overline{QR}$. By the Pythagorean theorem on $\triangle CDE$,
\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]
+179 Let $O_1$ and $O_2$ be the centers of circles $P$ and $Q$ respectively. Let $A$ and $B$ be the points where line $PQ$ intersects the circles $P$ and $Q$ respectively.
Let $C$ be the midpoint of $\overline{AB}$. Let $D$ be the foot of the perpendicular from $O_2$ to $\overline{PQ}$, and let $E$ be the foot of the perpendicular from $O_1$ to $\overline{O_2D}$.
Since $\overline{O_1E}$ and $\overline{O_2D}$ are both perpendicular to $\overline{PQ}$, $O_1EDO_2$ is a rectangle. Therefore, $O_1E=O_2D$.
Since $O_1O_2=12+8=20$, we know $O_1E=O_2D=10$. Because $\angle O_1AQ=\angle O_2BQ=90^\circ$, angles $O_1EA$ and $O_2DE$ are both right angles.
Thus, by the Pythagorean Theorem on $\triangle O_1AE$, \[AE=\sqrt{12^2-10^2}=2\sqrt{7}\]
and by the Pythagorean Theorem on $\triangle O_2DE$, \[ED=\sqrt{20^2-10^2}=10\sqrt3.\]
Finally, we find the length of $\overline{QR}$. By the Pythagorean theorem on $\triangle CDE$,
\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]
