From a point 10cm away from …
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Sia ? 6 years, 4 months ago
Let O be the centre of the given circle.
Then, OP = 10 cm. Also, PT = 8 cm.
Join OT.
Now, PT is a tangent at T and OT is the radius through the point of contact T.
{tex}\therefore \quad O T \perp P T {/tex}
In the right {tex}\Delta O T P {/tex}
we have
OP2 = OT2 + PT2 [by Pythagoras' theorem]
{tex}\Rightarrow{/tex} OT = {tex}\sqrt { O P ^ { 2 } - P T ^ { 2 } } = \sqrt { ( 10 ) ^ { 2 } - ( 8 ) ^ { 2 } } \mathrm { cm } = \sqrt { 36 } \mathrm { cm } {/tex} = 6cm.
Hence, the radius of the circle is 6 cm.
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