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PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at an external point T. Find the length of PT.

Answers (1)

Given radius $OP=OQ=5$

length of chord, $PQ=4\; cm$

$OT\perp PQ$

In right $\bigtriangleup O\!P\!M$ ,

$O\!P^2=P\!M^2+O\!M^2$

$(5)^2=(4)^2+O\!M^2$

$O\!M^2=25-16$

$O\!M=3\; cm$

In right $\bigtriangleup PT\!M$ ,

$PT^2=T\!M^2+P\!M^2 \;\;\;\;\;\;-(1)$

$\angle O\!PT=90^{\circ}$

$OT^2=PT^2+O\!P^2 \;\;\;\;\;\;-(2)$

From eqn (1) and (2), we get

$\Rightarrow OT^2=(T\!M^2+P\!M^2)+O\!P^2$

$\Rightarrow (T\!M+O\!M)^2=(T\!M^2+P\!M^2)+O\!P^2$

$\Rightarrow T\!M^2+O\!M^2+2\times T\!M \times O\!M=T\!M^2+P\!M^2+O\!P^2$

$\Rightarrow O\!M^2+2\times T\!M \times O\!M=P\!M^2+O\!P^2$

$\Rightarrow 3^2+2\times T\!M \times 3=4^2+5^2$

$\Rightarrow 9+6 T\!M =25+16$

$\Rightarrow 6 T\!M =32$

$\Rightarrow T\!M =\frac{16}{3}$

By eq (1) ,

$PT^2=T\!M^2+P\!M^2$

$=\left ( \frac{16}{3} \right )^2 + 4^2$

$= \frac{256}{9} + 16$

$= \frac{256+144}{9}$

$= \frac{400}{9}$

$PT^2=\left ( \frac{20}{3} \right )^2$

$PT= \frac{20}{3} \; cm$

The length of tangent PT is $\frac{20}{3} \; cm$

Posted by

Ravindra Pindel

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