how to prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$ How can one prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$?
I know that if i put: $x=y=0$,  i will get: $(0-a)^2+(0-b)=a^2+b^2=a^2+b^2$
But that's not a proof but checking.
Thanks. 

Notice, the equation $(x-a)^2+(y-b)^2=a^2+b^2$ represents a circle with a radius $\sqrt{a^2+b^2}$ & center at the point $(a, b)$. 
Now, the distance of the given point $(0, 0)$ from the center $(a, b)$ of the circle $$=\sqrt{(a-0)^2+(b-0)^2}$$ $$=\sqrt{a^2+b^2}$$  $$=\text{radius of circle}$$
But the above result is true only when the point $(0, 0)$ lies on the circumference of the circle i.e. the circle $(x-a)^2+(y-b)^2=a^2+b^2$ passes through the point $(0, 0)$
