Question

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

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

Answer 1

The correct option is B

$4(\sqrt{2}–1)$

Explanation for correct options

Let the radius of circle with least area be $r$,

Then, coordinates of centre$=(0,4-r)$.

Since, circle touches the line $y=x$ in first quadrant.

As perpendicuar distance is

The perpendicular distance $d$ of a line $Ax+By+C=0$ from a point $(x_1,y_1)$ is given by

$\mathit{d}\mathbf{=}\frac{\mathbf{|}\mathbf{A}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{b}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}\mathbf{c}\mathbf{|}}{\sqrt{\mathbf{(}{\mathbf{A}}^{\mathbf{2}}\mathbf{+}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}}}$

So by the figure $Ax+By+C$is $x-y=0$ and the point $=(0,4-r)$

$\begin{array}{rcl}& \Rightarrow & \left|\frac{0-(4-r)}{\sqrt{2}}\right|=r\\ & & \end{array}$

$\begin{array}{rcl}& \Rightarrow & r-4=\pm \sqrt{2}r\\ & & \end{array}$

$\begin{array}{rcl}& \Rightarrow & r\pm \sqrt{2}r=4\\ & & \end{array}$

$\begin{array}{rcl}\Rightarrow r(1\pm \sqrt{2})& =& 4\\ & & \end{array}$

$\begin{array}{rcl}& \Rightarrow & r=\frac{4}{1\pm \sqrt{2}}\end{array}$

So,

$r=\frac{4}{1+\sqrt{2}}$or $r=\frac{4}{1-\sqrt{2}}$

since $r\ne \frac{4}{1-\sqrt{2}}$as $\frac{4}{1-\sqrt{2}}<0$

as radius can't be less then zero

Hence the radius of the circle is $\frac{4}{1+\sqrt{2}}$.