Conics - Circle: Equation, Standard Form, and Graphing

A circle is one of the four conic sections, along with the parabola, the ellipse, and the hyperbola. A circle is the set of all points in a plane that are the same distance, called the radius, from one fixed point, called the center. That single idea, "same distance from a fixed point," is all you need to build the equation of a circle from scratch.

If a circle has center \((h, k)\) and radius \(r\), its standard form equation is:

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

This form is useful because you can read the center and radius straight off the equation. For example, in \((x - 3)^2 + (y + 2)^2 = 25\), the center is \((3, -2)\) and the radius is \(\sqrt{25} = 5\). Watch the signs carefully: the equation subtracts the coordinates of the center, so \((y + 2)^2\) really means \((y - (-2))^2\).

A circle centered at the origin simplifies to \(x^2 + y^2 = r^2\), since \(h = 0\) and \(k = 0\).

The standard form equation is really just the distance formula in disguise. If \((x, y)\) is any point on the circle and \((h, k)\) is the center, the distance between them must always equal the radius \(r\):

\(\sqrt{(x - h)^2 + (y - k)^2} = r\)

Squaring both sides removes the square root and gives the standard form equation \((x - h)^2 + (y - k)^2 = r^2\). This is why the radius always appears squared on the right side of the equation.

If instead you are given the two endpoints of a diameter rather than the center itself, use the midpoint formula to find the center first, then use the distance formula from that center to either endpoint to find the radius.

To graph \((x - h)^2 + (y - k)^2 = r^2\), plot the center \((h, k)\), then move a distance \(r\) up, down, left, and right from it to mark four points on the circle. Sketch the curve through those points. The figure below shows a circle with center \((2, 1)\) and radius \(4\), drawn as its upper and lower halves.

Notice that a single equation like \(y = 1 + \sqrt{16 - (x - 2)^2}\) only describes the top half, or semicircle. You need both the positive and negative square root branches together to get the full circle, since a circle is not the graph of a single function.

A circle equation is not always given in standard form. It often appears expanded, called the general form:

\(x^2 + y^2 + Dx + Ey + F = 0\)

To find the center and radius from general form, group the \(x\)-terms and \(y\)-terms and complete the square on each.

Convert \(x^2 + y^2 - 6x + 4y - 12 = 0\) to standard form.

Group like terms: \((x^2 - 6x) + (y^2 + 4y) = 12\)

Complete the square on each group. Half of \(-6\) is \(-3\), and \((-3)^2 = 9\). Half of \(4\) is \(2\), and \((2)^2 = 4\). Add both values to each side:

\((x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4\)

\((x - 3)^2 + (y + 2)^2 = 25\)

The center is \((3, -2)\) and the radius is \(5\), since \(\sqrt{25} = 5\).

A semicircle equation is simply half of a circle's standard form equation, solved for \(y\). Starting from \((x - h)^2 + (y - k)^2 = r^2\), isolate \(y\):

\(y = k \pm \sqrt{r^2 - (x - h)^2}\)

Choosing the \(+\) sign gives the upper semicircle, and choosing the \(-\) sign gives the lower semicircle. This is exactly the split you saw in the two graphs above.

Always check whether the terms are added or subtracted before reading off the center: an equation like \((x + 5)^2 + (y - 1)^2 = 9\) has center \((-5, 1)\), not \((5, 1)\). Also remember that the number on the right side of a standard form equation is \(r^2\), not \(r\), so take a square root to get the actual radius.

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