watch5826c

2022-06-24

Given a point $\left({x}_{0},{y}_{0}\right)$ and a radius r, how do you find the set of all circles that have that radius that pass through the point?

lorienoldf7

Step 1
We are interested in circles in the $⟨x,y⟩$ plane, therefore they will have Cartesian equation:
$\left(x-{x}_{C}{\right)}^{2}+\left(y-{y}_{C}{\right)}^{2}={r}^{2}$
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 $\left({x}_{P},\phantom{\rule{thinmathspace}{0ex}}{y}_{P}\right)$ and at the same time we have radius $r>0$ , it's evident that $\left({x}_{C},\phantom{\rule{thinmathspace}{0ex}}{y}_{C}\right)$ must belong to the circle with center $\left({x}_{P},\phantom{\rule{thinmathspace}{0ex}}{y}_{P}\right)$ and radius $r>0$ , that is:
$\left\{\begin{array}{l}{x}_{C}={x}_{P}+r\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(u\right)\\ {y}_{C}={y}_{P}+r\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(u\right)\end{array}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{with}\phantom{\rule{thickmathspace}{0ex}}u\in \left[0,\phantom{\rule{thinmathspace}{0ex}}2\pi \right)\phantom{\rule{thinmathspace}{0ex}}.$
In this way, we have determined the Cartesian equation of the sheaf of circles obtainable according to the chosen value of $u\in \left[0,\phantom{\rule{thinmathspace}{0ex}}2\pi \right)$

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