Find the equation of the circle tangent to the $x$-axis at the origin and tangent to the line $4x-3y+24=0$.

My Approach:

Let the equation of the required circle be: $$x^2+y^2+2gx+2fy+c=0$$

Let the equation to the tangent at origin $(0,0)$ to the above circle be $gx+fy+c=0$.

Then, what should I do? please help me to continue.

Thanks in Advance.

  • $\begingroup$ The wording is confusing (at least to me); Does "touching the $\;x\,-$ axis at the origin" = the circle's tangential to the $\;x\,-$ axis at the origin? $\endgroup$ – DonAntonio Oct 8 '16 at 11:39

A circle tangent to $x$-axis in $(0,0)$ has its center in $C_0(0,y_0)$ and its radius is $|y_0|$. Thus its equation has the form:


The distance from the center $C_0(x_0,y_0)=(0,y_0)$ of the circle to the straight line is (see (http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html)):


This distance must be equal to the radius $|y_0|$ giving the following equation:

$$\frac{|-3y_0+24|}{5}=|y_0| \ \ \Leftrightarrow \ \ -3y_0+24=\pm 5 y_0$$

which has 2 solutions: $y_0=3$ or $y_0=-12$ giving, (see (1)):

  • either the equation $x^2+(y-3)^2=3^2$ i.e., $x^2+y^2-6y=0$ (center $(0,3)$, radius 3),

  • or the equation $x^2+(y+12)^2=12^2$ i.e., $x^2+y^2+24y=0$ (center $(0,-12)$, radius = 12).

enter image description here

  • $\begingroup$ Hmm, how can this approach be modified to give the second solution? $\endgroup$ – Ian Miller Oct 8 '16 at 11:32
  • $\begingroup$ @JeanMarie, The question has two equations: $x^2+y^2+24y=0$ and $x^2+y^2-6y=0$. What do you have to say about that? $\endgroup$ – pi-π Oct 8 '16 at 11:34
  • $\begingroup$ You are right. I correct it $\endgroup$ – Jean Marie Oct 8 '16 at 11:34
  • $\begingroup$ Where did you get the two equations? $\endgroup$ – pi-π Oct 8 '16 at 11:38
  • $\begingroup$ Ah! The inclusion of the absolute value gives the second solution. Nicely done @JeanMarie. $\endgroup$ – Ian Miller Oct 8 '16 at 11:41

As the circle touches the x-axis at (0,0) then the centre of the circle must be on the y-axis and the y coordinate of the centre must be equal to the radius. So the centre is $(0,r)$. So the equation is:


Then you want to find the point of intersection of the circle and the straight line. Lets rearrange the straight line:


Sub in:


Multiple by $4^2$ to get rid of the fractions:




Collect and sort by powers of $y$:

$$25y^2-16(2r+9)y +576=0$$

Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero.


Pull out the $16^2$ term and factorise $576$ in preparation to divide out common factor.


Divide by common factor of $256$.


Expand and multiply:




Divide by 4:





So $r=3$ or $r=-12$

Note that $r=-12$ is a valid solution. It means that the circle is below the axis where earlier we assumed it was above the axis (when we set the centre to $(0,r)$).

So the two solutions are: $x^2+(y-3)^2=3^2$ and $x^2+(y+12)^2=12^2$

Or in expanded form: $x^2+y^2-6y=0$ and $x^2+y^2+24y=0$

enter image description here

  • 1
    $\begingroup$ I like this, its first principles and gets all the answers. $\endgroup$ – user24142 Oct 8 '16 at 11:36
  • 1
    $\begingroup$ There are many ways to approach this problem. I set out a technique which I thought would be suitable for your current level of knowledge based on the small start you made. $\endgroup$ – Ian Miller Oct 8 '16 at 11:38
  • $\begingroup$ What is meant by this: " Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero."? $\endgroup$ – pi-π Oct 8 '16 at 16:07
  • $\begingroup$ As the line is a tangent there is only one point of intersection with the circle. Secondly if a quadratic only has one solution then the discriminant is equal to zero $\Delta=b^2-4ac=0$ $\endgroup$ – Ian Miller Oct 8 '16 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.