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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.

Community
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Au101
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  • If the circle touches the y axis, then the roots are essentially and necessarily equal, – Someone May 11 '15 at 18:28
  • Write in standard form: $(x+g)^2 + (y+f)^2 = -c + g^2+f^2.$ "Touching" the $y$ is ambiguous, so I will assume this means "tangent" to the $y$-axis. The center of your circle is $(-g,-f).$ It must pass through the point $(0,-f)$ (point of tangency). This implies that radius satisfies $r^2=g^2.$ – matt biesecker May 11 '15 at 18:30
  • Your solution has exactly the right idea, but the way of concluding isn't quite right. In order to factorize as you did, you make the assumption $f^2 = \sqrt c$ which is what you want to prove (and more). – GPerez May 11 '15 at 18:36

2 Answers2

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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?

GPerez
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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 $f^2=c$.

wythagoras
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