The equation of the circle passing through the point (1,1) and having two diameters

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Text Solution

`x^(2)+y^(2)-4x-2y+5=0``x^(2)+y^(2)+4x+2y-5=0``x^(2)+y^(2)-4x-2y-5=0`none of these

Answer : C

Solution : The combined equation of two diameters is <br> `x^(2)-y^(2)-4x+2y+3=0` <br> `rArr (x-2)^(2)-(y-1)^(2)=0rArr (x+y-3)(x-y-1)=0` <br> Thus, the two diameters are x+y-3=0 and x-y-1=0 . <br> These two intersect at (2, 1). <br> Thus, the required circle has its centre at (2, 1) and passes through (-1, 2). So, its equation is <br> `(x-2)^(2)+(y-1)^(2)=(2+1)^(2)+(1-2)^(2)` <br> `rArr x^(2)+y^(2)-4x-2y-5=0`