Question

Let $C_1$ and $C_2$ denote the centres of the circles $x^2+y^2=4$ and $(x-2{)}^2+y^2=1$ respectively and let $P$ and $Q$ be their points of intersection. Then, the areas of $△C_{1}PQ$ and $C_{2}PQ$ are in the ratio

• A. $3:1$
• B. $5:1$
• C. $7:1$
• D. $9:1$

Answer 1

Answer: (C)

$7:1$

$x^2+y^2=4$ ...... (i)
$(x-2{)}^2+y^2=1$ ..... (ii)
$C_1=(0,0),C_2=(2,0)$

On solving Eqs. (i) and (ii), we get
$x=\frac{7}{4},y=\pm \frac{\sqrt{15}}{4}$

$C_{1}N=\frac{7}{4},PQ=\frac{2\sqrt{15}}{4}=\frac{\sqrt{15}}{2}$

$\frac{Area\left(△C_{1}PQ\right)}{Area\left(△C_{2}PQ\right)}=\frac{\frac{1}{2}\cdot PQ\cdot C_{1}N}{\frac{1}{2}\cdot PQ\cdot C_{2}N}$

$=\frac{C_{1}N}{C_{2}N}=\frac{\frac{7}{4}}{2-\frac{7}{4}}=\frac{\frac{7}{4}}{\frac{1}{4}}=7:1$

Hence, the required option is (c).