Problem:
Points A(6,13) and B(12,11) lie on 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:
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 (xβ6)2+132β=(xβ12)2+112β. 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:
2170βr=d40β, where r is the radius of the circle and d is the distance between the circle's center and (5,0). Therefore, d=17βr.
Using the Pythagorean Theorem on the right triangle OAX (or OBX ), we find that 170+r2=17r2, so r2=885β, and thus the area of the circle is (C)885βΟβ.
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=3xβ15.
Line AC is y=13xβ65. Therefore, the slope of the line perpendicular to AC is β131β, so its equation is y=β13xβ+13175β.
But notice that this line must pass through A(6,13) and (x,3xβ15). Hence 3xβ15=β13xβ+13175ββx=437β. So the center of the circle is (437β,451β).
Finally, the distance between the center, (437β,451β), and point A is 4170ββ. Thus the area of the circle is (C)885βΟβ
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 β³AOC is similar to β³DAC, so ACOAβ=DCADβ. The slope of AB is 6β1213β11β=3β1β, so the slope of CD is 3 . Hence, the equation of CD is yβ12=3(xβ9)βy=3xβ15. Letting y=0, we have x=5, so C=(5,0).
Now, we compute AC=(6β5)2+(13β0)2β=170β,AD=(6β9)2+(13β12)2β=10β, and DC=(9β5)2+(12β0)2β=160β.
Therefore OA=DCACβ ADβ=885ββ, and consequently, the area of the circle is Οβ OA2=(C)885βΟβ.
