Problem:
Points A(6,13) and B(12,11) lie on circle ω in the plane. Suppose that the tangent lines to ω at A and B intersect at a point on the x-axis. What is the area of ω ?
Answer Choices:
A. 883π​
B. 221π​
C. 885π​
D. 443π​
E. 887π​ Solution:
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is (x,0), the Pythagorean Theorem gives (x−6)2+132​=(x−12)2+112​. This simplifies to x=5.
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) AOBX is cyclic.
Therefore, we can apply Ptolemy's Theorem to give:
2170​r=d40​, where r is the radius of the circle and d is the distance between the circle's center and (5,0). Therefore, d=17​r.
Using the Pythagorean Theorem on the right triangle OAX (or OBX ), we find that 170+r2=17r2, so r2=885​, and thus the area of the circle is (C)885​π​.
OR
We firstly obtain x=5 as in Solution 1 . Label the point (5,0) as C. The midpoint M of segment AB is (9,12). Notice that the center of the circle must lie on the line passing through the points C and M. Thus, the center of the circle lies on the line y=3x−15.
Line AC is y=13x−65. Therefore, the slope of the line perpendicular to AC is −131​, so its equation is y=−13x​+13175​.
But notice that this line must pass through A(6,13) and (x,3x−15). Hence 3x−15=−13x​+13175​⇒x=437​. So the center of the circle is (437​,451​).
Finally, the distance between the center, (437​,451​), and point A is 4170​​. Thus the area of the circle is (C)885​π​
OR
The midpoint of AB is D(9,12). Let the tangent lines at A and B intersect at C(a,0) on the x-axis. Then CD is the perpendicular bisector of AB. Let the center of the circle be O. Then △AOC is similar to △DAC, so ACOA​=DCAD​. The slope of AB is 6−1213−11​=3−1​, so the slope of CD is 3 . Hence, the equation of CD is y−12=3(x−9)⇒y=3x−15. Letting y=0, we have x=5, so C=(5,0).
Now, we compute AC=(6−5)2+(13−0)2​=170​,AD=(6−9)2+(13−12)2​=10​, and DC=(9−5)2+(12−0)2​=160​.
Therefore OA=DCAC⋅AD​=885​​, and consequently, the area of the circle is π⋅OA2=(C)885​π​.
