Find all equations of a circle given two points

Question

Find all the equations of circles $a (x^2 + y^2) + b x + c y + d = 0$ through two given points, $(-1, 2)$ and $(3, 1)$.

I don't know how to approach this, I have to set up a matrix and solve but I'm not sure what to do.

Answer 1

Without loss of generality you may assume $a=1$

Plug your given points in $$ x^2+y^2 +bx +cy +d=0$$ to get two equations. $$5-b+2c+d=0$$ and $$10+3b+c+d=0$$

Thus we get $$b-2c-d=5$$ and $$ 3b+c+d=-10$$

Solve for $b$ and $c$ in terms of $d$ and plug in your equation.

My solution is $$7(x^2+y^2) -(d+15)x-(4d+25)y+d=0$$

Answer 2

It is simpler to use $(x-a)^2+(y-b)^2=r^2$ Get the line perpendicular to the segment connection the two points and bisection the segment. All the centers $(a,b)$ will lie on this line and for each center the radius ($r$) can be computed by using the distance between one of the given points and the center.

Answer 3

The circle centers lie on the line bisecting the two points, $y = 4 x - 5/2$, with a radius equal to the distance from the center to either of the points:

Choose an arbitrary $x$ then the equation of the circle is:

$$x^2 + (y - (4 x - 5/2))^2 = (x+3)^2 + ((4 x - 5/2)-1)^2$$

Answer 4

Consider the circle $f(x,y)=(x+1)(x-3)+(y-2)(y-1)=0$ that has the two points as endpoints of a diameter and the line $g(x,y)=x+4y-7=0$ through the two points. The solution set of every linear combination $s f(x,y)+t g(x,y)=0$ of these two equations also includes these two points. When $s=0$ and $t\ne0$, this is just the same line as before, while with $s\ne0$ this is an equation of a circle. In fact, all circles through the two points have an equation of this form. W.l.o.g. set $s=1$, expand the resulting equation and regroup to put the equation into the required form.

As an alternative not mentioned in other answers yet, there are a couple of ways to produce an equation of for a circle through the two given points and an arbitrary third point not colinear with them, but the resulting equation obscures the fact that there’s only one degree of freedom in the coefficients of the equations of this family of circles instead of two (the coordinates of the arbitrary point).

Answer 5

If you have two such equations, $f(x, y) = 0$ and $g(x, y) = 0$, then any affine combination of them, $$ (1-s) f(x, y) + s g(x, y) $$ will also be a circle passing through those two points (although this requires a little proof). And if $f$ and $g$ are distinct circles, then as $s$ ranges over the reals, you'll get all possible circles.

Finding one such equation isn't too hard: Use $$ P = \frac12 (-1, 2) + \frac12 (3, 1) = (1, \frac32) $$ as the center and the distance from $P$ to $(3, 1)$, namely $\sqrt{2^2 + \frac12^2} = \frac{3}{\sqrt{2}}$ as the radius, so you get $$ (x-1)^2 + (y-\frac32)^2 - \frac{9}{2} = 0 $$ as your first equation, i.e., $$f(x, y) = (x-1)^2 + (y-\frac32)^2 - \frac{9}{2}. $$ For the second, we can take the line between the two points, i.e, $$ 8x - 2y - 5 = 0, $$ so that $$ g(x, y) = 8x - 2y - 5. $$ Now $(1-s)f(x, y) + s g(x, y) = 0$ becomes \begin{align} (1-s)\left((x-1)^2 + (y-\frac32)^2 - \frac{9}{2}\right) + s (8x - 2y - 5) = 0 \end{align} You can expand that out to get coefficients for $a, b, c, d,$ etc.

Why does using a line for $g$ work? Because it's a representation of a circle of infinite radius. :) [It helps to have a little projective geometry to make sense of this.]