Question

$P$ is a point on the circle $C_1:q^2(x^2+y^2)=a^2{p}^2$

$Q$ is a point on the circle $C_2:x^2+y^2=a^2$

If the point $(p,q)$ lies on the line $y=2x,$ then $\frac{radiusof{C}_3}{radiusof{C}_2}$ equal to :

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{3}{2}$

Let $Q\left(x_1{y}_1\right)$ be a point on $C_2.$ Let $(m,n)$ be the point which divides $PQ$ is the ratio $p:q,$ then
$m=\frac{hq+x_{1}p}{p+q}$ and $n=\frac{kq+y_{1}p}{p+q}$
$x_1=\frac{m(p+q)-hq}{p}$ and $y_1=\frac{n(p+q)-kq}{p}$
Since $Q\left(x_1{y}_1\right)$ lies on $x^2+y^2=a^2$ we have $\left(m\right(p+q)-hq{)}^2+(n(p+q)-kq{)}^2=a^2{p}^2$
${(m-\frac{hq}{p+q})}^2+{(n-\frac{kq}{p+q})}^2=\frac{a^2{p}^2}{(p+q{)}^2}$
Locus of $(m,n)$ is
${(x-\frac{hq}{p+q})}^2+{(y-\frac{kq}{p+q})}^2=\frac{a^2{p}^2}{(p+q{)}^2}$
Which is circle $C_3$ if $(\alpha ,\beta )$ denote the centre of $C_3$, then
$\alpha =\frac{hq}{p+q}$ and $\beta =\frac{kq}{p+q}$
$h=(p+q)\frac{\alpha }{q}$ and $k=(p+q)\frac{\beta }{q}$
Since $P(h,k)$ lies on $C_1$
$q^2{\left(\frac{p+q}{q}\right)}^2({\alpha }^2+{\beta }^2)=p^2{a}^2$
${\alpha }^2+{\beta }^2=\frac{p^2{a}^2}{(p+q{)}^2}$
Locus of $(\alpha ,\beta )$ is therefore,
$x^2+y^2=\frac{p^2{a}^2}{(p+q{)}^2}$
which is the circle $C_4$ concentric with the circles $C_1$ and $C_2$ and radius of $C_4$ is same as that of $C_3$ .
Area of $C_4$. and $C_1$ are not same as their radii are not equal
$\frac{radiusof{C}_3}{radiusof{C}_2}$

$=1+\frac{p}{q}=\frac{3}{2}$
as $(p,q)$ lies on $y=2x$