Given a point $(x_0,y_0)$ and a radius $r$, how do you find the set of all circles that have that radius that pass through the point?

Let $(h,k)$ represent the center of the set of circles. Then, it's clear that we have the following, because we want the set of points $(h,k)$ that have a distance $r$ from $(x_0,y_0)$ such that the circle centered at $(h,k)$ has a radius of $r$:

$$ (x-h)^2 + (y-k)^2 = r^2$$

$$ (h-x_0)^2 + (k-y_0)^2 = r^2$$

It's unclear how to proceed from here. While it's true that we have two variables in two equations, and we must solve for $h$ and $k$, the algebra is really messy and I'm not entirely sure this problem admits a closed form solution. Any insights?


1 Answer 1


We are interested in circles in the $\langle x,y\rangle$ plane, therefore they will have Cartesian equation:

$$ (x-x_C)^2 + (y-y_C)^2 = r^2 $$

where their radius $r>0$ is fixed, so it remains to understand how to determine the centers.

In particular, given that we want the circles in question all to pass through $(x_P,\,y_P)$ and at the same time we have radius $r > 0$, it's evident that $(x_C,\,y_C)$ must belong to the circle with center $(x_P,\,y_P)$ and radius $r > 0$, that is:

$$ \begin{cases} x_C = x_P + r\,\cos(u) \\ y_C = y_P + r\,\sin(u) \\ \end{cases} \quad \quad \text{with} \; u \in [0,\,2\pi)\,. $$

In this way, we have determined the Cartesian equation of the sheaf of circles obtainable according to the chosen value of $u \in [0,\,2\pi)$; below a simple simulation in Mathematica:

{xP, yP, r, du} = {1, 2, 3, 2 Pi / 10};
{xC, yC} = {xP, yP} + r {Cos[u], Sin[u]};
circles = Table[(x - xC)^2 + (y - yC)^2 == r^2, {u, 0, 2 Pi - du, du}];
max = Max[Abs[{xP - 2 r, xP + 2 r, yP - 2 r, yP + 2 r}]];
ContourPlot[Evaluate[circles], {x, -max, max}, {y, -max, max},
            Axes -> True, AxesLabel -> {"x", "y"},
            Frame -> False, GridLines -> Automatic]

$\quad\quad\quad\quad\quad$enter image description here

  • $\begingroup$ A few questions, if you don't mind: $$$$ 1. Why the emphasis on "bundle"? Can there not be a set of circles? $$$$ 2. Wolfram Alpha does actually show a closed form for my system of equations when solving for $h$ and $k$, or $x_C$ and $y_C$ as you put it. What would happen if you solved for those variables and plugged them back into the first equation? Would you be able to see every circle in the set (or "bundle") at once? $\endgroup$
    – O.S.
    Mar 14, 2022 at 10:33

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