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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 isA. `x^(2)+y^(2)-6x-4y+4=0`B. `x^(2)+y^(2)-6x-4y-3=0`C. `x^(2)+y^(2)-6x-4y-4=0`D. `x^(2)+y^(2)-6x-4y+3=0`

Answer» Correct Answer - A
Two diameters `x+y=5` and `x-y=1` intersect at (3, 2) which is the centre of the circle. Let r be the radius of the circle. Then,
Area `=9 pi rArr pi r^(2)= 9 pi rArr r = 3`.
So, the equation of the circle is
`(x-3)^(2)+(y-2)^(2)=3^(2)` or, `x^(2)+y^(2)-6x-4y+4=0`.


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