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
26
B
46
C
215
D
10
E
233

Solution

## The correct option is D $$2\sqrt{15}$$From the given figure we can observe that $$TB^2=(TC)(TA)$$It is given that $$TC=6$$ and $$TA=TC+CA=10+6=16$$, therefore, $$TB^2=(TC)(TA)$$$$\Rightarrow TB^{ 2 }=6\times 16=96\\ \Rightarrow TB=\sqrt { 96 } =2\sqrt { 6 }$$$$\angle ACB$$ is inscribed in a semicircle and therefore, $$\angle ACB$$ is a right angle. Consequently $$\angle TCB$$ and $$\triangle TCB$$ is a right triangle.Using the pythagorean theorem:$$TC^{ 2 }+CB^{ 2 }=TB^{ 2 }\\ \Rightarrow (6)^{ 2 }+CB^{ 2 }=(2\sqrt { 6 } )^{ 2 }\\ \Rightarrow 36+CB^{ 2 }=96\\ \Rightarrow CB^{ 2 }=96-36\\ \Rightarrow CB^{ 2 }=60\\ \Rightarrow CB=\sqrt { 60 } =2\sqrt { 15 }$$ Maths

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