JEE MAIN - Mathematics (2004 - No. 45)
Intercept on the line y = x by the circle $${x^2}\, + \,{y^2} - 2x = 0$$ is AB. Equation of the circle on AB as a diameter is :
$$\,{x^2}\, + \,{y^2} + \,x\, - \,y\,\, = 0$$
$$\,{x^2}\, + \,{y^2} - \,x\, + \,y\,\, = 0$$
$$\,{x^2}\, + \,{y^2} + \,x\, + \,y\,\, = 0$$
$$\,{x^2}\, + \,{y^2} - \,x\, - \,y\,\, = 0$$
Explanation
Solving $$y=x$$ and the circle
$${x^2} + {y^2} - 2x = 0,$$ we get
$$x = 0,y = 0$$ and $$x=1,$$ $$y=1$$
$$\therefore$$ Extremities of diameter of the required circle are
$$\left( {0,0} \right)$$ and $$\left( {1,1} \right)$$. Hence, the equation of circle is
$$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0$$
$$ \Rightarrow {x^2} + {y^2} - x - y = 0$$
$${x^2} + {y^2} - 2x = 0,$$ we get
$$x = 0,y = 0$$ and $$x=1,$$ $$y=1$$
$$\therefore$$ Extremities of diameter of the required circle are
$$\left( {0,0} \right)$$ and $$\left( {1,1} \right)$$. Hence, the equation of circle is
$$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0$$
$$ \Rightarrow {x^2} + {y^2} - x - y = 0$$
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