Answer:
Note that the tangents to the circles at A and B intersect at a point Z on XY by radical center. Then, since∠ZAB=∠ZQA and ∠ZBA=∠ZQB. ∠AZB+∠AQB=∠AZB+∠ZAB+∠ZBA=180°. ∴ZAQB is cyclic.But if O is the center of w, clearly ZAOB is cyclic with diameter ZO, so ∠ZQO is 90° ⇒Q is the mid-point of XY.Then, by Power of a Point, PY· PX = PA · PB = 15 and it is given that PY+PX = 11. Thus,PX=(11±[tex]\sqrt{61}[/tex])/2. So, PQ=[tex]\frac{\sqrt{61} }{2}[/tex], PQ²=[tex]\frac{61}{4}[/tex]. Thus, the answer is 61+4=65.
Step-by-step explanation:
