Question

The radius of a circle, having minimum area, which touches the curve $y=4-x^2$ and the lines, $y=|x|$ is :

• A. $2(\sqrt{2}-1)$
• B. $4(\sqrt{2}-1)$
• C. $4(\sqrt{2}+1)$
• D. $2(\sqrt{2}+1)$

Answer 1

Answer: (B)

$x^2=-(y-4)$
Let a point on the parabola $P\left(\frac{t}{2},4-\frac{t^2}{4}\right)$
Equation of normal at P is
$y+\frac{t^2}{4}-4=\frac{1}{t}\left(x-\frac{t}{2}\right)$
$\Rightarrow x-ty-\frac{t^3}{4}+\frac{7}{2}t=0$
It passes through centre of circle, say (0, k)
$-tk-\frac{t^3}{4}+\frac{7}{2}t=\mathrm{0...}\left(i\right)$
$t=0,t^2=14-4k$
Radius$=r=\left|\frac{0-k}{\sqrt{2}}\right|$ (Length of perpendicular from (0, k) to y = x)
$\Rightarrow r=\frac{k}{\sqrt{2}}$
Equation of circle is $x^2+{\left(y-k\right)}^2=\frac{k^2}{2}$
It passes through point P
$\frac{t^2}{4}+{\left(4-\frac{t^2}{4}-k\right)}^{{}^2}=\frac{k^2}{2}$
$t^4+t^2\left(8k-28\right)+8k^2-128k+256=\mathrm{0...}\left(ii\right)$
For t = $0\Rightarrow k^2-16k+32=0$
$k=8\pm 4\sqrt{2}$
$\therefore r=\frac{k}{\sqrt{2}}=4(\sqrt{2}-1)$
(discarding $4(\sqrt{2}+1))\dots$ (iii)
For $t=\pm \sqrt{14-4k}$
${\left(14-4k\right)}^2+\left(14-4k\right)\left(8k-28\right)+8k^2-128k+256=0$
$2k^2+4k-15=0$
$k=\frac{-2\pm \sqrt{34}}{2}$
$\therefore r=\frac{k}{\sqrt{2}}=\frac{\sqrt{17}-\sqrt{2}}{2}$ (Ignoring negative ...(iv) value of r)
From (iii) & (iv),
$r_{min}=\frac{\sqrt{17}-\sqrt{2}}{2}$
But from options, $r=4\left(\sqrt{2}-1\right)$