Problem:
Points A(6,13) and B(12,11) lie on a 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:
Let T be the point where the tangents at A and B intersect. By symmetry T lies on the perpendicular bisector ℓ of AB, so in fact T is the (unique) intersection point of line ℓ with the x-axis. Computing the midpoint M of AB gives (9,12), and computing the slope of AB gives 6−1213−11​=−31​. This means that the slope of ℓ is 3, so the equation of ℓ is given by y−12=3(x−9). Setting y=0 yields that T=(5,0).
Now let O be the center of circle ω. Note that OA⊥AT and OB⊥BT, so in fact M is the foot of the altitude from A to the hypotenuse of △OAT. By the distance formula, TM=410​ and MA=10​. Then by the Altitude on Hypotenuse Theorem, MO=41​10​, so by the Pythagorean Theorem radius AO of circle ω is 41​170​. As a result, the area of the circle is
161​⋅170⋅π=(C)885π​​
OR
As above, the line y−12=3(x−9) passes through T(5,0) and the center of the circle O. The slope of line AT is 13 , so the slope of AO is −131​. The equation of line OA is 13(y−13)=6−x. Thus the intersection of line AO and line OT is O(437​,451​). Then the radius of the circle is
