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# Point Y is on a circle and point P lies outside the circle

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Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 10, then what is AB?

Nov 4, 2020

#1
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By Power of A Point, $$PB \cdot PA = PY^2$$.

Therefore, $$PB = \frac{PY^2}{PA} = \frac{10^2}{15} = \frac{20}{3}.$$

And as a result $$AB = PA - PB$$, meaning that $$AB$$ is $$15 - \frac{20}{3} = \frac{25}{3}$$

Nov 4, 2020
#2
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Important: If one secant and one tangent are drawn to a circle from one exterior point, then the square of the length of the tangent is equal to the product of the external secant segment and the total length of the secant.

PY2 = PB * PA

Nov 4, 2020