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Question

# Find the length of the tangent drawn to a circle with radius 8 cm from a point 17 cm away from the centre of the circle.

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Solution

## $\text{Let}\mathit{\text{O}}\text{be the centre of the given circle}\text{.}\phantom{\rule{0ex}{0ex}}\text{Let}\mathit{\text{P}}\text{be a point, such that}\phantom{\rule{0ex}{0ex}}\mathit{\text{OP}}=17\text{cm}\text{.}\phantom{\rule{0ex}{0ex}}\text{Let}\mathit{\text{OT}}\text{be the radius, where}\phantom{\rule{0ex}{0ex}}\mathit{\text{OT}}=5\text{cm}\phantom{\rule{0ex}{0ex}}\text{Join}\mathit{\text{TP}}\text{,where}\mathit{\text{TP}}\text{is a tangent}\text{.}\phantom{\rule{0ex}{0ex}}\text{Now,tangent drawn from an external point is perpendicular to the radius}\phantom{\rule{0ex}{0ex}}\text{at the point of contact}\text{.}\phantom{\rule{0ex}{0ex}}\therefore \text{OT}\perp \text{PT}\phantom{\rule{0ex}{0ex}}\text{In the right}△\mathit{\text{OTP}}\text{, we have:}\phantom{\rule{0ex}{0ex}}{\mathit{\text{OP}}}^{2}={\mathit{\text{OT}}}^{2}+{\mathit{\text{TP}}}^{2}\text{}\left[\text{By Pythagoras' theorem:}\right]\phantom{\rule{0ex}{0ex}}\mathit{\text{TP}}=\sqrt{{\mathit{\text{OP}}}^{2}-O{T}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{17}^{2}-{8}^{2}}\phantom{\rule{0ex}{0ex}}\text{=}\sqrt{289-64}\phantom{\rule{0ex}{0ex}}\text{=}\sqrt{225}\phantom{\rule{0ex}{0ex}}\text{=15 cm}\text{.}\phantom{\rule{0ex}{0ex}}\therefore \text{The length of the tangent is 15 cm}\text{.}$

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