Question

A circle of radius 1 unit touches the positive $X$-axis and positive $Y$-axis at $P$ and A variable line $L$ passing through the origin intersects the circle in two points $M$ an slope of the line L for which the area of the triangle MNQ is maximum, then fi $2010\left(m^2\right)$

Answer 1

Answer: 670

Clearly, equation of the circle is
$\Rightarrow (x-1{)}^2+(y-1{)}^2=1$
$\Rightarrow x^2+y^2-2x-2y+1=0$
Let the equation of the line be $y=mx$
$\Rightarrow mx-y=0$
So, $p=\left|\frac{m-1}{\sqrt{1+m^2}}\right|$ (where $p$ is the length of the perpendicular from $C$ on $MN$ )
Now, $l^2=1-p^2=1-\frac{(m-1{)}^2}{1+m^2}$
$\Rightarrow l^2=\frac{2m}{1+m^2}\Rightarrow l=\frac{\sqrt{2m}}{\sqrt{1+m^2}}\Rightarrow 2l=MN=\frac{2\sqrt{2m}}{\sqrt{1+m^2}}$
Perpendicular from $(0,1)$ on $mx-y=0$, is
$q=\left|\frac{-1}{\sqrt{1+m^2}}\right|=\frac{1}{\sqrt{1+m^2}}$
Hence area of the $\Delta QPN$ is $=\left(\frac{1}{2}\cdot 2l\cdot q\right)$
$A=\frac{1}{2}\frac{2\sqrt{2m}}{\sqrt{1+m^2}}\cdot \left(\frac{1}{\sqrt{1+m^2}}\right)$
$A=\frac{\sqrt{2m}}{\left(1+m^2\right)}=f(m)\text{ (say) }$

Hence $f^{\prime }(m)=\sqrt{2}\left[\frac{\frac{\left(1+m^2\right)}{2\sqrt{m}}-\sqrt{m}\cdot 2m}{{\left(1+m^2\right)}^2}\right]=0$ or $\left(1+m^2\right)-4m^2=0$
Also, $3m^2=1\Rightarrow m=\frac{1}{\sqrt{3}}$.
Hence, $2010\left(m^2\right)=670$