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Equation of a Circle: Center, Radius and Completing the Square

Equation of a Circle: Center, Radius and Completing the Square

Written by Jean Carlos Novaes

A circle is the set of points in a plane at a fixed positive distance from a center. That distance is the radius. Its equation lets you test whether a point lies on the curve and recover the center and radius even when they are hidden in expanded algebra.

Standard form shows the center directly

Let the center be C(a, b) and radius r. The horizontal and vertical differences between C and a point (x, y) form the legs of a right triangle. The distance formula therefore gives the standard equation below.

(x−a)2+(y−b)2=r2

The center coordinates have the opposite signs to the constants inside the squared brackets. A center at the origin gives the special case with no shifts. A diameter passes through the center and has length twice the radius; other chords need not pass through it.

The center C is displaced from the origin. A radius connects C to a point on the circle, and its horizontal and vertical components form a right triangle.
The center C is displaced from the origin. A radius connects C to a point on the circle, and its horizontal and vertical components form a right triangle.

General form and the radius condition

Expanding the standard form gives equal coefficients for the squared coordinate terms, no product of x and y, and two linear terms. Using coefficients D, E and F, the center is obtained by negating half of D and half of E. The radius squared is shown separately.

x2+y2+Dx+Ey+F=0
r2=D2+E24−F

The resulting radius squared must be positive for a nondegenerate real circle. Zero gives a single point; a negative value gives no real points. If both squared terms share another nonzero coefficient, divide the whole equation by it first.

Worked example: recover the center by completing squares

Group the x terms and the y terms. Half of −6 is −3, so completing the first square requires adding 9. Half of 4 is 2, so the second needs 4. Move the constant to the other side and add those same amounts there to preserve equality.

x2+y2−6x+4y−12=0
(x−3)2+(y+2)2=25
C=(3,−2),r=5

The final equation has center (3, −2) and radius 5. Check by expanding the squared expressions: their constants sum to 13, and subtracting 25 restores the original constant −12. Reading 25 as the radius would confuse the radius with its square.

Circumference length and enclosed area

The circle as a curve has a circumference length; its enclosed disk has an area. Both depend on the radius, but they describe different measurements and use different units.

C=2πr
A=πr2

The length has linear units and the area square units. Retaining π gives an exact expression; decimal values are approximations. The unit circle is the special case centered at the origin with radius 1.

References

  • OpenStax. Intermediate Algebra 2e: Distance and Midpoint Formulas; Circles.
Jean Carlos Novaes

About Jean Carlos Novaes

I hold a degree in Computer Science from the Federal University of Bahia (2017) and I am the editor and founder of this website.

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