Circle locus from a fixed point and distance
In analytic geometry, a locus is a set of points that satisfy a condition. A circle is one of the most
important loci: it is the set of all points that are the same distance from a fixed point called the center.
If the center is \(C(h,k)\) and the distance is \(r\), then every point \(P(x,y)\) on the locus satisfies
\(CP=r\). This is exactly how a compass draws a circle: you fix the needle at the center and keep the pencil tip
at a constant radius.
Using the distance formula, the distance from \(P(x,y)\) to the center \(C(h,k)\) is
\[
CP=\sqrt{(x-h)^2+(y-k)^2}.
\]
The circle condition \(CP=r\) becomes
\[
\sqrt{(x-h)^2+(y-k)^2}=r.
\]
Squaring both sides (which is valid when \(r\ge 0\)) gives the standard center–radius equation of a circle:
\[
(x-h)^2+(y-k)^2=r^2.
\]
This form is usually the best for understanding geometry directly: you can immediately read the center \((h,k)\)
and the radius \(r\).
Sometimes you need the circle written in an expanded “general” form for algebraic work:
\[
x^2+y^2+Dx+Ey+F=0.
\]
Expanding \((x-h)^2\) and \((y-k)^2\) and collecting like terms produces
\[
x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0,
\]
so \(D=-2h\), \(E=-2k\), and \(F=h^2+k^2-r^2\). Both equations describe the same locus; they are just different
ways to present it.
There is one important special case: if \(r=0\), the “circle” collapses to a single point (the center). The equation
still works:
\[
(x-h)^2+(y-k)^2=0
\]
is satisfied only by \(x=h\) and \(y=k\). In other words, the locus contains exactly one point.
Common mistakes include forgetting to square the radius (the equation uses \(r^2\), not \(r\)), mixing up the signs
in \((x-h)\) and \((y-k)\), or allowing a negative radius. This tool enforces \(r\ge 0\) and displays results in a
clear center–radius form and an expanded form so you can match whatever format your homework or textbook expects.
