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The intercept on the line `y=x` by the circle `x^2+y^2-2x=0` is AB. Equation of the circle with AB as a diameter is (A) `(x-1/2)^3+(y-1/2)^2=1/2` (B) `(x-1/2)^2+(y-1/2)^2=1/4` (C) `(x+1/2)^2+(y+1/2)^2=1/2` (D) `(x+1/2)^2+(y+1/2)^2=1/4` |
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Answer» Correct Answer - `x^(2)+y^(2)-x-y=0` Equation of any circle passing through the point of lt brgt intersection of `x^(2)+y^(2)-2x=0 and y=x`is `(x^(2)+y^(2)-2x)+lambda(y-x)=0` `rArr x^(2)+y^(2)-(2+lambda)x+lambday=0` Its centre is `((2+lambda)/(2),(-lambda)/(2))`. `rArr` For AB to be the diameter of the required circle the centre must be on AB, i.e. `2+lambda=-lambda` `rArr lambda = -1 " "[therefore y = x]` therefore, equation of the required circle is `x^(2)+y^(2)=(2-1)x-1*y=0` `rArr x^(2)+y^(2)-x-y=0` |
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