Question

A variable line $\frac{x}{a}+\frac{y}{b}=1$ moves such that, the harmonic mean of $a^2$ and $b^2$ is $32.$ If the locus of the foot of the perpendicular from the origin to the above line is a circle with radius ${}^{\prime }{r}^{\prime },$ then the value of $4r^2+1$ is

Answer 1

H.M. of $a^2$ and $b^2=\frac{2a^2{b}^2}{a^2+b^2}=32$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{16}$
Variable line is $\frac{x}{a}+\frac{y}{b}=1...\left(1\right)$
Any line perpendicular to above line and passing through origin is $\frac{x}{b}-\frac{y}{a}=0...\left(2\right)$

Let $P(\alpha ,\beta )$ be the foot of perpendicular from the origin.
$P$ lies on both $\left(1\right)$ and $\left(2\right)$
$\Rightarrow \frac{\alpha }{a}+\frac{\beta }{b}=1...\left(3\right)$
and $\frac{\alpha }{b}-\frac{\beta }{a}=0...\left(4\right)$

$(3{)}^2+(4{)}^2$
$\Rightarrow {\alpha }^2(\frac{1}{a^2}+\frac{1}{b^2})+{\beta }^2(\frac{1}{b^2}+\frac{1}{a^2})=1$
$\Rightarrow {\alpha }^2+{\beta }^2=16$

$\therefore$ Locus of $P$ is $x^2+y^2=16$
$\Rightarrow$ Radius of the circle, $r=4$
$\therefore 4r^2+1=65$