Question

If $(\alpha ,\beta )$ is a point on the circle whose centre is on the x-axis and which touches the line $x+y=0$ at $(2,-2)$, then find the greatest integral value of ${}^{\prime }{\alpha }^{\prime }$.

Answer 1

Equation of circle is $(x-h{)}^2+(y-k{)}^2=r^2$

A center of circle lies on $x-axis,$

Hence, $k=0$

$\Rightarrow$ $(x-h{)}^2+y^2=r^2$ ------ ( 1 )

Now,

Slope of line $x+y=2$

$\Rightarrow$ $m=-\frac{coefficientofx}{coefficientofy}=-1$

Differentiating ( 1 ) w.r.t. $x$ we get,

$2(x-h)+2y\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-2(x-h)}{2y}=\frac{-(x-h)}{y}$

$\Rightarrow$ ${\frac{dy}{dx}}_{(2,-2)}=\frac{-(2-h)}{-2}=\frac{2-h}{2}$

$\Rightarrow$ $\frac{2-h}{2}=-1$

$\Rightarrow$ $2-h=-2$

$\Rightarrow$ $h=4$

$\Rightarrow$ $(x-4{)}^2+y^2=r^2$

Now, $(2,-2)$ lies on circle.

$\Rightarrow$ $(2-4{)}^2+(-2{)}^2=r^2$

$\Rightarrow$ $4+4=r^2$

$\Rightarrow$ $r^2=8$

$\Rightarrow$ $r=2\sqrt{2}$

$\Rightarrow$ $(x-4{)}^2+y^2=(2\sqrt{2}{)}^2$

Any general point on this circle will be $(4+2\sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta )$

$\Rightarrow$ $(\alpha ,\beta )$ $=(4+2\sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta )$

$\Rightarrow$ $\alpha =4+2\sqrt{2}\cos \theta$

Now, maximum value of $\cos \theta =1$

$\Rightarrow$ ${\alpha }_{max}=4+2\sqrt{2}$

Hence, greatest inetegral value of $\alpha$ will be $4+2\sqrt{2}$