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 the greatest value of $\alpha$ is

• A. $4-\sqrt{2}$
• B. $6$
• C. $4+2\sqrt{2}$
• D. $4+\sqrt{2}$

Answer 1

The correct option is C $4+2\sqrt{2}$

Consider the given figure which explains the above condition.
Triangle $OAB$ is a right angled isosceles triangle right angled at B.
Hence
$OA=2\sqrt{2}units$.
And $\angle AOB={45}^0$
Now
Triangle $OAC$ is also a right angled triangle right angled at A and $\angle AOB={45}^0$
Hence $OA=AC=2\sqrt{2}units$
Hence by Pythagoras theorem $OC=4units.$
Hence
$h=4units.$
Hence equation of the circle is
$(x-4{)}^2+y^2=A{C}^2$
$(x-4{)}^2+y^2=8$
Now
$x-4=2\sqrt{2}\cos \theta$
$x=4+2\sqrt{2}\cos \theta$
$y=2\sqrt{2}\sin \theta$
For $\theta =0$
$x=4+2\sqrt{2}$
Which has been denoted by E.