Question

The equation of the circle having $x-y-2=0$ and $x-y+2=0$ as two tangents and $x-y=0$ as diameter is

• A. $x^2+y^2+2x-2y+1=0$
• B. $x^2+y^2-2x+2y-1=0$
• C. $x^2+y^2=2$
• D. $x^2+y^2=1$

Answer 1

Answer: (C)

$x^2+y^2=2$

Since, the equations of tangents $x-y-2=0$ and $x-y+2=0$ are parallel.
Therefore, distance between them $=$ Diameter of the circle
$=\frac{|2-(-2)|}{\sqrt{1^2+1^2}}$ $(\because \frac{C_2-C_1}{\sqrt{a^2+b^2}})$
$=\frac{4}{\sqrt{2}}=2\sqrt{2}$
Hence, radius $=\frac{1}{2}\left(2\sqrt{2}\right)=\sqrt{2}$
It is clear from the figure that centre lies on the origin.
Therefore, equation of circle is ${(x-0)}^2+{(y-0)}^2={\left(\sqrt{2}\right)}^2$.
$\Rightarrow x^2+y^2=2$