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

Answer 1

Answer: (B)

Since the equations of tangents x - y - 2 = 0 and x — y + 2 = 0 are parallel.

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