MHT CET · Maths · Circle
The equation of the circle passing through the point \((1,1)\) and having two diameters along the pair of lines \(x^2-y^2-2 x+4 y-3=0\) is
- A \((x+2)^2+(y-2)^2=4\)
- B \((x-3)^2+(y-1)^2=4\)
- C \((x-1)^2+(y-2)^2=1\)
- D \((x+1)^2+(y+2)^2=1\)
Answer & Solution
Correct Answer
(C) \((x-1)^2+(y-2)^2=1\)
Step-by-step Solution
Detailed explanation
Center \((h,k)\): \(2x-2=0 \Rightarrow x=1\), \(-2y+4=0 \Rightarrow y=2\). So, \((1,2)\). Radius squared \((r^2)\): \(r^2 = (1-1)^2 + (1-2)^2 = 0^2 + (-1)^2 = 1\).
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