# Find the equation of a circle which is tangent to $y$-axis at a given point and cuts a chord of given length on the $x$-axis

How to find the equation of the circle which touches $y$ axis at $(0,3)$ and cuts a chord of length $8$ on the $x$ axis?

It should look like this:

### My approach:

Since the circle touches $y$ axis at $(0,3)$, its center has $y$-coordinate $3$. So the equation of the circle is of the form $(x-r)^2+(y-3)^2=r^2$.

How can I proceed further by using the fact that the circle passes through $(a,0)$ and $(b,0)$ with $b-a=8$? Anyway this would be very long. Is there some alternative?

• What does "cuts 8 intercepts" mean? – TonyK Jun 29 '14 at 15:42
• @TonyK.. I have added the picture – Rob Jun 29 '14 at 15:49
• Given the points you have marked, it is clear by symmetry that the centre is on the line $x=5$ – Mark Bennet Jun 29 '14 at 15:49
• @MarkBennet But without the graph, how would I proceed? – Rob Jun 29 '14 at 15:58
• Apparently you want to say the circle "cuts a chord of length 8 on the $x$ axis", rather than the "cuts 8 intercepts" formulation. – hardmath Jun 29 '14 at 16:05

HINT:

For $x$ intercept $y=0$

Setting $y=0, x^2-2xr+9=0$

If the circle cuts $x$ axis at $(x_1,0);(x_2,0)$ where $x_1\ge x_2$

We have $x_1-x_2=8$ and $\displaystyle x_1+x_2=2r,x_1x_2=9$

Use $\displaystyle (x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$ to find $r$

Suppose you have just the points $(0,3)$ where the circle is tangent, and a horizontal chord of length $8$ along the $x$-axis. There are two basic positions or the circle - to the left or to the right of the $y$-axis. Assume to the right (to the left is simply a reflection).

Construct the vertical line through the centre of the circle: this bisects the chord of length $8$ along the $x$-axis, and draw the radius to the point you have labelled $(1,0)$ [but whose co-ordinates we don't yet know]. You have a right-angled triangle whose legs have lengths $3$ vertically and $4$ horizontally, so the hypotenuse, which is the radius, has length $5$.

Let $\Omega(x_\Omega,y_\Omega)$ be the center of your circle. The vertical axis is tangent to the circle at $D(0,3)$, means that $\Omega$ is on the horizontal line $y=3$. So $y_\Omega=3$. Now, $A(0,1)$ and $B(0,9)$ are on the circle, so $\Omega$ is on the bisector of the line segment from $A$ to $B$, which is the verticle line $x=5$. Hence, $x_\omega=5$.

Therefore, $\Omega(5,3)$ and $\Omega D$ is a radius so the equation of the circle is $$(x-5)^2+(y-3)^2=25$$

$\begin{vmatrix} x^2+y^2&x&y&1\\ 0^2+3^2&0&3&1\\ 1^2+0^2&1&0&1\\ 9^2+0^2&9&0&1\\ \end{vmatrix} = \begin{vmatrix} x^2+y^2&x&y&1\\ 9&0&3&1\\ 1&1&0&1\\ 81&9&0&1\\ \end{vmatrix}= 24(x^2+y^2)-240x-144y+216= x^2+y^2-10x-6y+9=0$