Two circles are externally tangent at T. The line AB is a common external tangent to the two circles, and P is the foot of the altitude from T to line AB. Find the length TP.

faiafronk Mar 17, 2024

#2**0 **

Since T is the point of external tangency, line OT is the radius of the larger circle, and line TC is the radius of the smaller circle.

Since ∠OTP and ∠TCP are right angles, triangle OTP and triangle TCP are right triangles.

We are given that OT=7 and TC=3. Since the altitude from T to line AB bisects the common external tangent, AP=PB=2AB.

By the Pythagorean Theorem on triangle OTP, we have OP2=OT2−TP2=72−TP2, and by the Pythagorean Theorem on triangle TCP, we have OP2=TC2+CP2=32+(AP−TP)2=9+(2AB−TP)2.

Setting these two expressions for OP2 equal, we get:

\begin{align*} 7^2 - TP^2 &= 9 + \left( \frac{AB}{2} - TP \right)^2 \ 49 - TP^2 &= 9 + \frac{AB^2}{4} - AB \cdot TP + TP^2 \ 0 &= \frac{AB^2}{4} - AB \cdot TP + 58 \end{align*}

Since we are looking for TP, let's rewrite this equation to isolate TP. We can factor the quadratic:

\begin{align*} 0 &= \frac{AB^2}{4} - AB \cdot TP + 58 \ 0 &= (AB - 14)(AB - 4) \end{align*}

Thus, either AB−14=0 or AB−4=0. Since a negative length for a line segment doesn't make sense, we must throw out the solution AB=14.

Therefore, AB=4, and applying the Pythagorean Theorem on triangle OTP one more time, we find that TP = sqrt(7^2 - (AB/2)^2) = sqrt(41).

Boseo Mar 19, 2024