Answer

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Hint: Use the information that tangent to a circle is perpendicular to the radius through the point of contact.

We know that, tangent to a circle is perpendicular to the radius through the point of contact. So, $\angle OTP = {90^ \circ }$. In right angle triangle OTP, we have

$

O{P^2} = O{T^2} + P{T^2} \\

\Rightarrow {13^2} = O{T^2} + {12^2} \\

\Rightarrow O{T^2} = {13^2} - {12^2} \\

\Rightarrow O{T^2} = 169 - 144 \\

\Rightarrow O{T^2} = 25 \\

\Rightarrow OT = 5 \\

$

Hence the radius of the circle is 5 cm.

Note: Using the correct theorem is the key to solve this problem. After that Pythagoras theorem will give the required result.

We know that, tangent to a circle is perpendicular to the radius through the point of contact. So, $\angle OTP = {90^ \circ }$. In right angle triangle OTP, we have

$

O{P^2} = O{T^2} + P{T^2} \\

\Rightarrow {13^2} = O{T^2} + {12^2} \\

\Rightarrow O{T^2} = {13^2} - {12^2} \\

\Rightarrow O{T^2} = 169 - 144 \\

\Rightarrow O{T^2} = 25 \\

\Rightarrow OT = 5 \\

$

Hence the radius of the circle is 5 cm.

Note: Using the correct theorem is the key to solve this problem. After that Pythagoras theorem will give the required result.

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