COMEDK · Maths · 12. Circle
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 \(x^{2}+y^{2}-2 x+2 y-1=0\)
- D \(x^{2}+y^{2}+2 x-2 y+1=0\)
Answer & Solution
Correct Answer
(B) \(x^{2}+y^{2}=2\)
Step-by-step Solution
Detailed explanation
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…
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