Question

A circle of radius $\sqrt{8}$ is passing through origin and the point $(4,0)$. If the centre lies on the line $y=x$, then the equation of the circle is

• A. $(x-2{)}^2+(y-2{)}^2=8$
• B. $(x+2{)}^2+(y+2{)}^2=8$
• C. $(x-3{)}^2+(y-3{)}^2=8$
• D. $(x+3{)}^2+(y+3{)}^2=8$
• E. $(x-4{)}^2+(y-4{)}^2=8$

Answer 1

The correct option is A $(x-2{)}^2+(y-2{)}^2=8$

Let the equation of the circle be $(x-a{)}^2+(y-a{)}^2=8[\because \text{Centre lies on}x=y\text{and radius is}\sqrt{8}]$

Both $(0,0)$ and $(4,0)$ lies of the circle, therefore we have the below 2 equations

$a^2+a^2=8⟹a=\pm 2$

$(4-a{)}^2+a^2=8⟹a^2-4a+4=0⟹(a-2{)}^2=0⟹a=2$

$\therefore ,a=2$ and the required equation is $(x-2{)}^2+(y-2{)}^2=8$