Question

The intercept of the line $y=x$ on the circle $x^2+y^2-2x=0$ is $AB$. Equation of circle with $AB$ as a diameter is given by

• A. $x^2+y^2+x+y=0$
• B. $x^2+y^2-x+y=0$
• C. $x^2+y^2-(x+y)=0$
• D. $x^2+y^2+x-y=0$

Answer 1

Answer: (C)

$x^2+y^2-(x+y)=0$

Given equation of circle

$(x-1{)}^2+y^2=1$

Intersection with $y=x$

$(x-1{)}^2+x^2=1$

$2x^2-2x+1=1$

$2x(x-1)=0$

$x=0,1y=0,1$

Diametric point $(0,0)(1,1)$

Centre of circle $=(\frac{1}{2},\frac{1}{2})$ and radius $=\frac{1}{\sqrt{2}}$

The equation of the circle with centre $(h,k)$ and the radius $a$ is,

$(x-h{)}^2+(y-k{)}^2=a^2$

Required equation of circle ${(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=\frac{1}{2}$

$x^2+y^2-x-y=0$