Help setting out a proof about the circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$

Question

16. Given that the circle

$$x^{2} + y^{2} + 2gx + 2fy + c = 0$$

touches the $y$-axis, prove that $f^{2} = c$.

So, because the circle touches the $y$-axis, we know that there is a solution to this equation where $x = 0$, so we can say:

$y^{2} + 2fy + c = 0$

And now the answer seems so attainable, because, when we factorise this quadratic, we will have:

$(y + f)^{2} = 0$, where $f = \sqrt{c}$. But I just don't know - is that enough to constitute the proof? It doesn't seem very well explained, I'm just going on my experience of quadratic equations.

I tried rewriting the equation as:

$(y + f)^{2} - f^{2} + c = 0$, which seems so close - but I haven't been able to do anything with this.

So, how do I do this formally.

Answer 1

By "touches" I take it the problem intends tangency, or "touches at exactly one point". If this is the case the answer can be gotten by the quadratic formula, or actually just its discriminant.

Now, call $(x_0,y_0)$ the point of tangency (it's nice to distinguish between particular points and "variables" that are used in the equation of the circle). Of course, $x_0 = 0$ like you said. You also correctly stated that $y_0$ must satisfy the quadratic $$y_0^2+2fy_0+c = 0$$ Look at the discriminant of the quadratic: $$\Delta = 4f^2 - 4c = 4(f^2 - c)$$ If $y_0$ is the only point where the circle intersects the $y$-axis, what must the discriminant be?

Answer 2

If the $y$-axis touches the circle, then the middle point lies $r$ above the $y$-axis so it has middle point $(r,a)$ or $(-r,a)$ for some $a$. Therefore the formula of the circle is $(x-r)^2+(y-a)^2=r^2$ or $(x+r)^2+(y-a)^2=r^2$

We can rewrite your equation as $(x+g)^2+(x+f)^2=g^2+f^2-c$. Further, $g=\pm r$, so $g^2=r^2$, so the equation is $(x+g)^2+(x+f)^2=r^2+f^2-c$. This gives $f^2-c=0$, so $f^2=c$.