Question 16 goes as follows:

16. Given that the circle

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

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

A circle, with its centre in the first quadrant, touches the $y$-axis and also touches externally the circle

$$x^{2} + y^{2} - 4x = 5;$$

prove that the coordinates of its centre satisfy the equation

$$y^{2} = 10x + 5.$$

If the circle also touches the $x$-axis, prove that the abscissa of the point of contact is $5 + \sqrt{30}$.

I have already completed the first part of this proof (with the help of this very website). I thought that once I had done that, I would be able to complete the rest of the question on my own, but it was not to be.

The first thing I can do is write down what I know.

I know, from what I've proved, that the equation of this circle can be expressed in the form:

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

I know that, since the circle touches the $y$-axis and has its centre in the first quadrant, the $x$-coordinate of the centre is $r$ (if $r$ is the radius of the circle).

I know that the radius of the circle $x^{2} + y^{2} - 4x = 5$ is 3 and that its centre is $(2, 0)$.

The first thing I tried to do with this information was to say: let the centre of the circle be $(r, d)$.


$(x - r)^{2} + (y - d)^{2} = r^{2} \Leftrightarrow x^{2} + y^{2} - 2rx - 2dy + d^{2} = 0$

Next I said, well, I know that there is a point $(a, b)$ that lies on both of my circles (their point of contact) and so:

$a^{2} + b^{2} - 2ra - 2db + d^{2} = 0$


$a^{2} + b^{2} - 4a - 5 = 0$

But all I've managed to do after all of this is give myself two simultaneous equations with four unknowns and I just can't for the life of me come up with any more useful information.

  • $\begingroup$ Perhaps you could construct a system, but I am not sure. $\endgroup$ – Jimmy360 May 12 '15 at 2:13

I am using your symbols and their interpretation. Let $(r,d)$ be the center of the circle you are trying to find. Since it touches externally, therefore the distance between the centers should be equal to the sum of the radii of the two circles. This means \begin{align*} (r-2)^2+(d-0)^2 & = (r+3)^2\\ r^2-4r+4+d^2 & = r^2+6r+9\\ d^2 & = 10r+5. \end{align*}

If the circle also touches the $x-$axis then $d=r$. In which case you have the quadratic equation $$r^2-10r-5=0.$$ It has two possible roots: $$r = \frac{10 \pm \sqrt{120}}{2}=5 \pm \sqrt{30}.$$ But $r>0$, hence $r=5 + \sqrt{30}$.

  • $\begingroup$ nice. you beat me to the answer.(+1) $\endgroup$ – abel May 12 '15 at 2:35
  • $\begingroup$ That does all seem very neat, thank you, but I'm not quite sure I get why the distance between the centers should be equal to the sum of the radii of the two circles. I follow all of your working after that. I've just been messing around with a piece of paper and a pencil and it "feels" true, but I'm not sure I've ever seen any proof that this must always be so $\endgroup$ – Au101 May 12 '15 at 2:43
  • $\begingroup$ As I understand it, what you're saying is true: the distance between the centers should be equal to the sum of the radii of the two circles, if the radii have the same gradient at the point of contact. That is, the two radii will form a straight line between the centres. Is that true? Because surely, if the radii met at an angle, the distance between the centres would not be the sum of the radii, you'd have to construct a triangle and find the length of the third side. So I take it we need to prove the radii form this straight line. This is because the circles are tangential to each other? $\endgroup$ – Au101 May 12 '15 at 2:55
  • 1
    $\begingroup$ @Au101 at the tangent point the tangent line drawn is perpendicular to the radius and since that will hold for both circles, therefore the radii of both have to be perpendicular to the (same) tangent line. This means the radii will be collinear. $\endgroup$ – Anurag A May 12 '15 at 6:05
  • $\begingroup$ Lovely stuff, now we're on the same page $\endgroup$ – Au101 May 12 '15 at 6:40

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