Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\overline{PA}$ and $\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9.
OA is a radius = 9
Because OAP is a right angle, OP = sqrt ( PA^2 + OA^2) = sqrt (12^2 + 9^2) =
sqrt (225) = 15
Let AB intersect OP at R
And triangles OPA and OAR are similar
So
PA / OP = AR / OA
12 / 15 = AR / 9
AR = 108/15 = 36/5
And AR = (1/2) AB ...so
AB = 2 (AR) = 2(36/5) = 72/5
Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\overline{PA}$ and $\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9.
see: https://web2.0calc.com/questions/circle-tangent