+0  
 
0
4
1
avatar+36 

The two circles below are externally tangent. A common external tangent intersects line PQ at R. Find QR

Circle radius-> 12 (p)and 8(q)

 Sep 14, 2024
 #1
avatar+544 
0

To solve this problem, we need to find the length \( QR \) given that the two circles are externally tangent and have radii of 8 and 12.

 

Let the centers of the two circles be denoted as \( O_1 \) and \( O_2 \), with \( O_1 \) being the center of the circle with radius 8 and \( O_2 \) being the center of the circle with radius 12. The distance between the centers \( O_1O_2 \) can be calculated as:

\[
O_1O_2 = r_1 + r_2 = 8 + 12 = 20.
\]

Next, we denote the lengths of the segments from the tangent point to the circle centers. Let \( d \) represent the distance between the circle centers, which we've calculated to be 20.

When a common external tangent is drawn, it intersects the line joining the centers of the two circles at a right angle at the point where it meets the external line \( PQ \). The formula for the length of the tangent \( t \) from an external point to a circle is given by:

\[
t = \sqrt{d^2 - (r_1 + r_2)^2},
\]


where \( d \) is the distance between the centers of the circles and \( r_1 \) and \( r_2 \) are the radii of the circles. However, here we use the tangent distance formula as follows:

\[
t^2 = d^2 - (r_1 - r_2)^2.
\]

Here, we have:

- \( r_1 = 8 \),
- \( r_2 = 12 \),
- \( d = 20 \).

Now we calculate \( r_2 - r_1 \):

\[
r_2 - r_1 = 12 - 8 = 4.
\]

Now substituting into the tangent formula:

\[
t^2 = d^2 - (r_2 - r_1)^2.
\]

Calculating \( d^2 \):

\[
d^2 = 20^2 = 400,
\]

and \( (r_2 - r_1)^2 \):

\[
(r_2 - r_1)^2 = 4^2 = 16.
\]

Substituting these values into the equation gives:

\[
t^2 = 400 - 16 = 384.
\]

Now, take the square root to find \( t \):

\[
t = \sqrt{384} = \sqrt{64 \times 6} = 8\sqrt{6}.
\]

Therefore, the length of the common tangent \( t \) is \( 8\sqrt{6} \).

Now, the problem requires finding \( QR \). Assuming that point \( R \) is the point where the tangent intersects line \( PQ \), and assuming line \( PQ \) is parallel to the line joining centers \( O_1 \) and \( O_2 \), we have that:

The distance \( QR \) from the tangent point to the line can often be represented relative to the tangent lengths, but specifics of \( QR \) in positions relative to other given points were not provided. If we are seeking \( QR \) as the length of the common external tangent, we state that:

\[
QR = 8\sqrt{6}.
\]

Thus, the final answer is:

\[
QR = 8\sqrt{6}.
\]

 Sep 14, 2024

3 Online Users

avatar
avatar