Question

From the point $P(16,7)$ tangent $PQ$ and $PR$ are drawn to the circle $x^2+y^2-2x-4y-20=0$. If $C$ be the centre of the circle then area of the quadriteral $PQCR$ is

• A. $450sq.units$
• B. $15sq.units$
• C. $50sq.units$
• D. $75sq.units$

Answer 1

The correct option is D $75sq.units$

Radius of circle$=\sqrt{1+4+20}=5$

$\therefore l\left(CP\right)=\sqrt{(1-16{)}^2+(2-7{)}^2}$

$=\sqrt{225+25}$

$\therefore l\left(CP\right)=5\sqrt{10}$

$\therefore l\left(PQ\right)=\sqrt{\left(l\right(CP){)}^2-l(CQ{)}^2}$

$\therefore l\left(PQ\right)=\sqrt{250-25}=15$

$\therefore$ Area of $\Delta CQP=\frac{1}{2}\times CQ\times QP$

$=\frac{1}{2}\times 5\times 15$

$\therefore$ Area of $\square$PQCR$=2\times$ Area of $\Delta$PQC

$=2\times \frac{1}{2}\times 5\times 15$

$=75$.