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

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

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