How to determine family of circles passing through two given points?

Question

The question asks to show that the equation of any circle passing through two given points takes a certain form. I have obtained the points as being $(2,1)$ and $(2,-1)$ but I'm not sure as to how to determine the general form of the family of circles that pass through these points from just this information.

For anyone that's interested, I've posted the entire question below.

Show that the equation of any circle passing through the points of intersection of the ellipse

$$(x+2)^2+2y^2=18$$

and the ellipse

$$9(x-1)^2+16y^2=25$$

can be written in the form

$$x^2-2ax+y^2=5-4a.$$

Thanks guys.

Answer 1

For the first one, if the center is $(h,k)$

We have $(2-h)^2+(1-k)^2=(2-h)^2+(1+k)^2$ (each being the radius$^2$)

$\implies4k=0\iff k=0$

So, the equation will be $(x-h)^2+(y-0)^2=(2-h)^2+1$

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The equation any two dimensional curve passing through the intersection of the given ellipses can be written as

$$(x+2)^2+2y^2-18+k[9(x-1)^2+16y^2-25]=0$$ where $k$ is an arbitrary constant

For a circle, the coefficients of $x^2,y^2$ must be same

Use this fact to determine the value of $k$

Answer 2

Based on your result, we know the center of the circle must be located on $x$-axis. Set the coordinate of center as $(a,0)$, then you can express radius of the circle in terms of $a$, which is $(a-2)^2+1$.

Then any point $(x,y)$ on the circle must satisfy $(x-a)^2+y^2=(a-2)^2+1$. Simplify it.

Answer 3

THIS ANSWER GIVES YOU THE FAMILY OF CIRCLES THROUGH TWO GIVEN POINTS.

The way I would work is something like this... if the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are on your circle then the line segment $[P_1P_2]$ is a chord. Suppose that the distance $|P_1P_2|=2d$.

Now the perpendicular from the centre of a circle to a chord bisects the chord. We can get the equation of this perpendicular bisector because we have that the midpoint of $[P_1P_2]$ is on it... $((x_1+x_2)/2,(y_1+y_2)/2)$. The slope of the perpendicular bisector, $m$, satisfies $$m\cdot m_{[P_1P_2]}=-1,$$ so is

$$m=-\frac{x_2-x_1}{y_2-y_1},$$

so the perpendicular bisector has equation

$$y=-\frac{x_2-x_1}{y_2-y_1}x+\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2},$$

briefly $k=mh+c$ where $k\sim y$, $h\sim x$ and

$$m=-\frac{x_2-x_1}{y_2-y_1},$$

and

$$c=\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2}$$

Here $h$ is free --- each $h$ gives you a different centre $(h,k)=(h,mh+c)$.

By drawing a picture you will see a RAT triangle with vertices at, say the midpoint of $[P_1P_2]$, the centre $(h,mh+c)$ and $P_1$. Using Pythagoras we have that

$$r^2=d^2+\left(\frac{x_1+x_2}{2}-h\right)^2+\left(\frac{y_1+y_2}{2}-(mh+c)\right)^2$$

So we have that the circles in question are:

$$\{(x-h)^2+(x-(mh+c))^2=r^2:h\in\mathbb{R}\},$$

where each of $m$ and $c$ are functions of $P_1,\,P_2$ and --- once $h$ is chosen --- $r$ is a function of $P_1$, $P_2$ and $h$.