how to prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$

Question

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.

Answer 1

The equation $$(x-x_M)^2+(y-y_M)^2=r^2$$ describes the circle (to be more specific the circle line) with center $(x_M,y_M)$ and radius $r$. This means, that every point $(a,b)$ that lies on the circle line fulfills this equation. Also the other way around, every point $(a,b)$ that fulfills this equation must lie on the circle line. In your situation you have $x_M=a,y_M=b$ and $r^2=a^2+b^2$.

So if you have to show that the circle passes through a specific point, it is indeed enough to just pluck the point in the given equation and see if everything turns out well.

Answer 2

I suppose this illustrates the difference between "showing" or "proving" as opposed to "verifying".

Just plugging in the values $x=0$ and $y=0$ and getting LHS=RHS is verifying the result.

To show or prove it you could show that the distance from the origin to the centre is the same as the radius, which is straightforward enough.

Answer 3

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 $(0, 0)$