+0  
 
0
25
1
avatar+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.

 Feb 12, 2024

Best Answer 

 #1
avatar+195 
0

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 $\triangle CDE$,

\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]

 Feb 13, 2024
 #1
avatar+195 
0
Best Answer

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 $\triangle CDE$,

\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]

Boseo Feb 13, 2024

2 Online Users

avatar