Two distinct points A and B are on a circle with center at O, and point P is outside the circle such that PA and PB are tangent to the circle. Find AB if PA = 15 and the radius of the circle is 9.
+130477 PA = PB because tangents drawn from an external point to a circle are equal
Triangle PAO is right with angle PAO = 90
So PO is the hypotenuse, leg PA = 15 and leg OA = the radius
So
PO = sqrt ( PA^2 + OA^2) = sqrt ( 15^2 + 9^2) = sqrt 306 = 3sqrt 34
The area of this triangle = (15 * 9) / 2 = 135/2
And PBOA will form a kite with an area of twice [ PAO ] = 135
And PO and AB will be the diagonals of this kite
And the area of this kite = product of diagonal lengths / 2
So
135 = 3sqrt 34 * AB / 2
270 = 3sqrt 34 * AB
AB = 270 / (3sqrt 34 ) = 90 / sqrt 34 units ≈ 15.43 units
