## How To Write The Equation of A Circle

### Equation Of A Circle

A circle is a cross sectional cut of a conic section, just as is a parabola, ellipse and hyperbola.Â A circle is characterized by its uniform radius and symmetry about the x and y axes.Â Â The standard equation of a circle is written in the following form.

Â

(x â€“ h)^{2} + (y â€“ k)^{2} = r^{2}

Â

The x and y coordinates represent any point that lies along the curve of the circle, h and k are the x and y coordinates, respectively of the center point of the circle and r is the radius of the circle.Â The circle is located at the intersection of the x and y axis (the origin), when h = k = 0 and at this point the equation becomes, x^{2} + y^{2} = r^{2}.Â Additionally, this equation can be rewritten as r = (x^{2} + y^{2})^{1/2}.

Â

Letâ€™s apply this concept to a few examples.

Â

1.) Find the coordinates of a possible point that lies along the curve of the circle, based on the following equation.

Â

(x â€“ 4)^{2} + (y â€“ 6)^{2} = 64

Â

We can see that the center of the circle is located at x = 4 and y = 6.Â We also know that

r^{2} = 64 and so r = 8.Â Since the radius is 8, we select a simple point Â Â Â Â Â to consider, 8 units to the right of the center point.Â This means that we are considering the point (4 + 8, 6 + 0) or (12, 6) on the curve.Â We can check make Â Â Â Â Â Â Â Â sure this point correctly lies along the curve by plugging in these values for the x Â and y coordinates and making sure our equation is equivalent to 64.

Â

At (12, 6) we have:

Â

(x â€“ 4)^{2} + (y â€“ 6)^{2} = 64

(12 â€“ 4)^{2} + (6 â€“ 6)^{2} = 64

(8)^{2} + (0)^{2} = 64

(8)^{2} = 64, Correct.

Â

2.) What is the radius of a circle centered at (2, 3) and has the point (-6, 9) lying on its curve.

Â

The equation of our circle has the form, (x â€“ h)^{2} + (y â€“ k)^{2} = r^{2}.Â And if we apply the values for the center point, we have (x â€“ 2)^{2} + (y â€“ 3)^{2} = r^{2}.Â Now we add the values for the point lying along the curve giving us (-6 â€“ 2)^{2} + (9 â€“ 3)^{2} = r^{2}.Â From here we can determine the radius of the circle.

Â

(-6 â€“ 2)^{2} + (9 â€“ 3)^{2} = r^{2}

(-8)^{2} + (6)^{2} = r^{2}

64 + 36 = r^{2}

100 = r^{2}

r = 10

Â

The radius of this circle is 10.

Â

Â

REFERENCE:

Bittinger, Marvin L., Ellenbogen, David, Keedy, Mervin L. (1994).Â Intermediate Algebra: Concepts and Application (4^{th} ed.).Â USA: Addison-Wesley Publishing Company, Inc.