# Finding the value of $k$ for the equation of a circle

I have been told that the circle with equation $x^2 + y^2 - 12x -10y + k=0$ meets the co-ordinate axes exactly three times, and I have to find the value of $k$.

Now, I first found the centre of the circle, with the information given, to be $(6,5)$, and substituing this into the equation, we obtain $k=61$. Yet, $k$ cannot equal $61$ since that would imply the radius of the circle is zero, a contradiction to the fact that the equation is a circle. From this, I concluded that $k=0$ (the answer in the marking instructions), yet the marking instructions does not state my solution (although, I do know it is not correct).

In the marking instructions, there are two solutions, $k=25$ and $k=0$, and they are found, respectively, by assuming that the circle is tangent to the y-axis and from this calculating the radius of the circle (which would then provide the value of $k$), or that the circle touches the origin and from this calculating the radius of the circle. Now, I don't know if their solutions are correct or not, because they don't exactly show that their obtained value of $k$ satisfies the condition on the circle (that it meets the co-ordinate axes exactly three times).

My questions are whether these solutions are the only solutions and and whether it's possible to show that they are indeed the only solutions.

• You need to transform the equation to the form $(x-a)^2+(y-b)^2=r^2$, where $a,b$ and $r$ will be expressions containing $k$. Only when that is done can you start analysing where the circle lies in the coordinate system for different $k$. – Arthur Jan 9 '16 at 12:28
• Sorry I meant to say $k$ cannot equal $61$, I edited it. – user303801 Jan 9 '16 at 12:30
• @user303801 you can't substitute the centre of the circle into the circle equation as the centre of the circle does not lie on the circle.... – J.Cork Jan 9 '16 at 12:43
• Sorry, a terrible mistake. – user303801 Jan 9 '16 at 12:59

A circle can meet an axis $0$, $1$ or $2$ times, so if it meets the two co-ordinate axes exactly three times in total then it must cross one axis at two points. The possibilities are:

• it is tangent to the $y$-axis, and crosses the $x$-axis twice
• it is tangent to the $x$-axis, and crosses the $y$-axis twice
• it passes through the point $(0,0)$, crosses the $x$-axis at another point and crosses the $y$-axis at a third point

If you consider each of these three possibilities then you will have a complete solution.

Completing the squares imply the centre of the circle is at $(6,5)$ so these cases become:

• possible, as having radius $6$ would satisfy the conditions
• impossible, as having radius $5$ to be a tangent would then not cross the $y$-axis
• possible, as having radius $\sqrt{6^2+5^2}$ would satisfy the conditions
• Thanks a lot. I appreciate it. – user303801 Jan 9 '16 at 12:48

You first need to bear in mind what the equation of a circle with centre $(a,b)$ and radius $r$ is. This is given by

\begin{align} (x-a)^2+(y-b)^2=r^2. \end{align}

Your equation is not quite in the form needed, so you need to complete the square in x and y to get

\begin{align} (x-6)^2+(y-5)^2 + k - 36 - 25 =0. \end{align}

As you correctly state, this is of centre $(6,5)$. You now need to use the fact it intersects with the coordinate axes exactly 3 times. Drawing the picture, you know this is only possible if the radius of the circle is equal to $6$ (so that it touches the $y$-axis and intersects the $x$-axis twice), or if it is equal to $\sqrt{61}$ (when the circle passes through the origin). This means you need to solve

$$k-36-25 = -36,$$

which gives $k=25$, and

$$k-36-25 = -61,$$

which gives $k=0$. So according to my calculations, the solutions are correct.

• Thanks a lot. I appreciate it. – user303801 Jan 9 '16 at 12:49

HINT:

If $$(x-6)^2+(y-5)^2=61-k$$ meets the co-ordinate axes exactly three times,

either exactly one of the axes will be tangent to the circle.

or one intersection is common i.e., $(0,0)$

Now if a straight line is tangent to a given circle $\iff$

the length of the radius will be equal to the perpendicular distance of the straight line from the center

• It seems that you are missing the case when $k=0$. – mathlove Jan 9 '16 at 12:49
• @mathlove, Rectified – lab bhattacharjee Jan 9 '16 at 13:07