Problem:
Points A(6,13) and B(12,11) lie on a circle Ο in the plane. Suppose that the tangent lines to Ο at A and B intersect at a point on the x-axis. What is the area of Ο?
Answer Choices:
A. 883Οβ
B. 221Οβ
C. 885Οβ
D. 443Οβ
E. 887Οβ
Solution:
Let T be the point where the tangents at A and B intersect. By symmetry T lies on the perpendicular bisector β of AB, so in fact T is the (unique) intersection point of line β with the x-axis. Computing the midpoint M of AB gives (9,12), and computing the slope of AB gives 6β1213β11β=β31β. This means that the slope of β is 3, so the equation of β is given by yβ12=3(xβ9). Setting y=0 yields that T=(5,0).
Now let O be the center of circle Ο. Note that OAβ₯AT and OBβ₯BT, so in fact M is the foot of the altitude from A to the hypotenuse of β³OAT. By the distance formula, TM=410β and MA=10β. Then by the Altitude on Hypotenuse Theorem, MO=41β10β, so by the Pythagorean Theorem radius AO of circle Ο is 41β170β. As a result, the area of the circle is
161ββ 170β Ο=(C)885Οββ
OR
As above, the line yβ12=3(xβ9) 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 β131β. The equation of line OA is 13(yβ13)=6βx. Thus the intersection of line AO and line OT is O(437β,451β). Then the radius of the circle is
