Question

Tangent $\overline{TB}$ and secant $\overline{TCA}$ are drawn to circle $O$. Diameter $\overline{AB}$ is drawn. If $TC=6$ and $CA=10,$ then $CB=$

• A. $2\sqrt{6}$
• B. $4\sqrt{6}$
• C. $2\sqrt{15}$
• D. $10$
• E. $2\sqrt{33}$

Answer 1

Answer: (C)

$2\sqrt{15}$

From the given figure we can observe that $T{B}^2=\left(TC\right)\left(TA\right)$

It is given that $TC=6$ and $TA=TC+CA=10+6=16$, therefore,

$T{B}^2=\left(TC\right)\left(TA\right)$

$\Rightarrow T{B}^2=6\times 16=96\Rightarrow TB=\sqrt{96}=2\sqrt{6}$

$\angle ACB$ is inscribed in a semicircle and therefore, $\angle ACB$ is a right angle.

Consequently $\angle TCB$ and $△TCB$ is a right triangle.

Using the pythagorean theorem:

$T{C}^2+C{B}^2=T{B}^2\Rightarrow (6{)}^2+C{B}^2=(2\sqrt{6}{)}^2\Rightarrow 36+C{B}^2=96\Rightarrow C{B}^2=96-36\Rightarrow C{B}^2=60\Rightarrow CB=\sqrt{60}=2\sqrt{15}$