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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 $$CB =$$

534531_32ef4cd147fa4ec88209738d6c73a907.png


A
26
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B
46
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C
215
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D
10
loader
E
233
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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:
$$TC^{ 2 }+CB^{ 2 }=TB^{ 2 }\\ \Rightarrow (6)^{ 2 }+CB^{ 2 }=(2\sqrt { 6 } )^{ 2 }\\ \Rightarrow 36+CB^{ 2 }=96\\ \Rightarrow CB^{ 2 }=96-36\\ \Rightarrow  CB^{ 2 }=60\\ \Rightarrow CB=\sqrt { 60 } =2\sqrt { 15 }$$ 

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