Question

Tangent to the circle $x^2+y^2$ = 4 at any point on it in the first quadrant makes intercepts OA and OB on x and y axes respectively, O being the centre of circle. Find the minimum value of (OA + OB).

Answer 1

Given circle $x^2+y^2=4$ ....(i) Let P $(\alpha ,\beta )$ $\therefore {\alpha }^2+{\beta }^2$ ....(ii)

Also $2x+2y\frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$ $\therefore {\begin{array}{c}\frac{dy}{dx}\end{array}|}_{\text{at P}}=-\frac{\alpha }{\beta }=m_r$

Equation of tangent : $y-\beta =-\frac{\alpha }{\beta }(x-a)$ $\Rightarrow \alpha x+\beta y=a^2+{\beta }^2$

By (ii), $\alpha x+\beta y=4$ $\Rightarrow \frac{x}{\frac{4}{\alpha }}+\frac{y}{\frac{4}{\beta }}=1$ $\therefore x-intercept=\frac{4}{\alpha }=OA,y-intercept=\frac{4}{\beta }=OB$

Let $OA+OB=s=\frac{4}{\alpha }+\frac{4}{\beta }\Rightarrow s=\frac{4}{\alpha }+\frac{4}{\sqrt{4-{\alpha }^2}}\Rightarrow \frac{ds}{d\alpha }=-\frac{4}{{\alpha }^2}+{\frac{4\alpha }{(4-{\alpha }^2)}}^{\frac{3}{2}}$

$\Rightarrow \frac{d^{2}s}{d{\alpha }^2}=\frac{8}{{\alpha }^3}+\frac{(4-{\alpha }^2{)}^{\frac{3}{2}}4-4\alpha \times \frac{3}{2}(4-{\alpha }^2{)}^{\frac{1}{2}}(-2\alpha )}{(4-{\alpha }^2{)}^3}\Rightarrow \frac{d^{2}s}{d{\alpha }^2}=\frac{8}{{\alpha }^3}+\frac{8({\alpha }^2+2)}{(4-{\alpha }^2{)}^{\frac{5}{2}}}$

For local maxima and/or minima, $\frac{ds}{d\alpha }=-\frac{4}{{\alpha }^2}+\frac{4\alpha }{(4-{\alpha }^2{)}^{\frac{3}{2}}}=0\Rightarrow {\alpha }^3=(4-{\alpha }^2{)}^{\frac{3}{2}}$

$\Rightarrow {\alpha }^6=(4-{\alpha }^2{)}^3=64-48{\alpha }^2+12{\alpha }^4-{\alpha }^6\Rightarrow m^3-6m^2+24m-32=0,$ where m = ${\alpha }^2$

$\Rightarrow (m-2)=0$ or $(m^2-4m+16)=0\therefore m=2,m^2-4m+16\ne 0\forall mϵR$

$\therefore \alpha =\sqrt{2}.[\therefore$ P lies in 1st quadrant] $\begin{array}{c}\therefore {\frac{d^{2}s}{d{\alpha }^2}}_{\text{at}\alpha =\sqrt{2}}\end{array}|=\frac{8}{2\sqrt{2}}+\frac{8(2+2)}{(4-2{)}^{\frac{5}{2}}}=6\sqrt{2}>0$

So, s is minimum at $\alpha =\sqrt{2}.$

Therefore, the minimum value of s =$\frac{4}{\sqrt{2}}+\frac{4}{\sqrt{4-2}}=4\sqrt{2}.$