Let PQ and RS be tangent at extremities of diameter PR of a circle of radius r . If PS and RQ i

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10-1.Circle and System of Circles

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Let $PQ$ and $RS$ be the tangent at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $(PQ.RS)$ is equal to

A

$(PX).(RX)$

B

$(QX).(SX)$

C

$(PX)^2 + (RX)^2$

D

$(QX)^2 + (SX)^2$

Solution

$\frac{{PQ}}{{RP}} = \frac{{PR}}{{RS}}$

$PQ.RS = {\left( {PR} \right)^2}$

$ = {\left( {PX} \right)^2} + {\left( {RX} \right)^2}$

Standard 11

Mathematics

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