Question

A circle of radius $\sqrt{5}$ units has diameter along the angle bisector of the lines $x+y=2$ and $x-y=2$. If chord of contact from the origin makes an angle of ${45}^{\circ }$ with the positive direction of $x$-axis, then the equation of the circle is

• A. $(x+2{)}^2+(y-2{)}^2=5$
• B. $(x-2{)}^2+(y+2{)}^2=5$
• C. $(x-2{)}^2+y^2=5$
• D. $(x+2{)}^2+y^2=5$

Answer 1

The correct option is B $(x-2{)}^2+(y+2{)}^2=5$

Equation of angle bisectors are $\frac{|x+y-2|}{\sqrt{1^2+1^2}}=\frac{|x-y-2|}{\sqrt{1^2+1^2}}$
i.e., $x=2$ and $y=0$
Now, centre cannot lie on $y=0$ because then chord of contact from the origin will always be parallel to $y-$axis.
So, let the centre be $(2,\alpha )$.
Then, equation of circle is
$(x-2{)}^2+(y-\alpha {)}^2=5$
$\Rightarrow x^2+y^2-4x-2\alpha y+{\alpha }^2-1=0$

Chord of contact, $T=0$ from $(0,0)$ is
$0\cdot x+0\cdot y-4\left(\frac{x+0}{2}\right)-2\alpha \left(\frac{y+0}{2}\right)+{\alpha }^2-1=0$
$\Rightarrow 2x+\alpha y-{\alpha }^2+1=0$
Slope $=\tan {45}^{\circ }=1$
$\Rightarrow -\frac{2}{\alpha }=1\Rightarrow \alpha =-2$
Hence, equation of the circle is $(x-2{)}^2+(y+2{)}^2=5.$