Question

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

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