Question:
If a circle has two of its diameters along the lines \(x+y=5\) and \(x-y=1\) and has area \(9\pi\), then the equation of the circle is :
If a circle has two of its diameters along the lines \(x+y=5\) and \(x-y=1\) and has area \(9\pi\), then the equation of the circle is :
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The centre is the intersection of the two diameters; the area gives the radius.
Updated On: Nov 12, 2026
- \(x^{2}+y^{2}-6x-4y-3=0\)
- \(x^{2}+y^{2}-6x-4y-4=0\)
- \(x^{2}+y^{2}-6x-4y+3=0\)
- \(x^{2}+y^{2}-6x-4y+4=0\)
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The Correct Option is D
Solution and Explanation
Diameters meet at the centre: adding \(x+y=5\) and \(x-y=1\) gives \(x=3\), then \(y=2\). Centre \((3,2)\).
Area \(\pi r^{2}=9\pi\Rightarrow r=3\).
Circle: \((x-3)^{2}+(y-2)^{2}=9\Rightarrow x^{2}+y^{2}-6x-4y+13-9=0\Rightarrow x^{2}+y^{2}-6x-4y+4=0\).
Answer: (4) \(x^{2}+y^{2}-6x-4y+4=0\)
Area \(\pi r^{2}=9\pi\Rightarrow r=3\).
Circle: \((x-3)^{2}+(y-2)^{2}=9\Rightarrow x^{2}+y^{2}-6x-4y+13-9=0\Rightarrow x^{2}+y^{2}-6x-4y+4=0\).
Answer: (4) \(x^{2}+y^{2}-6x-4y+4=0\)
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