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Point $Y$ is on a circle and point $P$ lies outside the circle such that $\overline{PY}$ is tangent to the circle. Point $A$ is on the circle such that segment $\overline{PA}$ meets the circle again at point $B$. If $PA = 15$ and $PY = 9$, then what is $AB$? [asy] unitsize(2 cm); pair A, B, P, Y; Y = dir(130); A = dir(20); B = dir(180); P = extension(Y, Y + rotate(90)*(Y), A, B); draw(Circle((0,0),1)); draw(Y--P--A); label("$A$", A, E); label("$B$", B, SW); label("$P$", P, W); label("$Y$", Y, N); [/asy]

1 Answer

9 votes

Answer:

48/5

Explanation:

By power of a point on P, we have PB * PA = PY^2. Therefore,

PB = PY^2/PA = 9^2/15 = 27/5. Then AB = PA - PB = 15 - 27/5 = 48/5

User Zulko
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