Question

$P$ lies in the exterior of $⨀(0,5)$ such that $OP=13$. Two tangents are drawn to the circle which touch the circle in $A$ and $B$. Find $AB$.

Answer 1

Given, center$(0,5)OP=13$

Radius of circle $=OB=5cm$

In $△OBP$

$(OB{)}^2+(BP{)}^2=(OP{)}^2{5}^2+(BP{)}^2=(13{)}^{2}BP=12cm$

Let $\angle OPB=\theta =\angle OPA$ then, $\sin \theta =\frac{OB}{OP}=\frac{5}{13}\left(1\right)$

Let $M$ be the midpoint of $AB$

$\angle OBM=\theta$

In $△OMB$

$\cos \theta =\frac{BM}{OB}BM=OB\cos \theta =5\sqrt{1-{\sin }^2\theta }$

From $\left(1\right)$

$BM=5\times \sqrt{1-{\left(\frac{5}{13}\right)}^2}=5\times \frac{12}{13}=\frac{60}{13}AB=2BM=\frac{120}{13}units$