Q85 / 180 · COMEDK 2016 · Mathematics

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
    $x^{2}+y^{2}=1$
  • Correct answer
    B
    $x^{2}+y^{2}=2$
  • C
    $x^{2}+y^{2}-2 x+2 y-1=0$
  • D
    $x^{2}+y^{2}+2 x-2 y+1=0$

Solution

Since, the equation of tangents $x-y-2=0$ and $x-y+2=0$ are parallel. Therefore, distance between them $=$ Diameter of the circle $\begin{aligned} &=\left|\frac{2-(-2)}{\sqrt{1^{2}+1^{2}}}\right| \\ &=\frac{4}{\sqrt{2}}=2 \sqrt{2} \end{aligned}$ Radius $=\frac{1}{2}(2 \sqrt{2})=\sqrt{2}$ Now, required equation of circle is $\begin{aligned} (x-0)^{2}+(y-0)^{2} &=(\sqrt{2})^{2} \\ x^{2}+y^{2} &=2 \end{aligned}$

Hence, the correct answer is option: 'B'.

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