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 Dependent Variable

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Equations of Circles

A circle can be defined as the set of all points in a plane that are equidistant from a fixed point. The fixed point is the center of the circle, and the distance between the center and a point on the circle is the radius (see the figure below).

You can use the Distance Formula to write an equation for the circle with center (h, k) and radius r. Let (x, y) be any point on the circle. Then the distance between (x, y) and the center (h, k) is given by

By squaring both sides of this equation, you obtain the standard form of the equation of a circle.

Standard Form of the Equation of a Circle

The point (x, y) lies on the circle of radius r and center (h, k) if and only if (x - h)2 + (y - k)2 = r2.

The standard form of the equation of a circle with center at the origin, (h, k) = (0, 0), is x2 + y2 = r2.

If r = 1, the circle is called the unit circle.

Example 1

Finding the Equation of a Circle

The point (3, 4) lies on a circle whose center is at (-1, 2), as shown in the figure below. Find an equation for the circle.

Solution

The radius of the circle is the distance between (-1, 2) and (3, 4).

You can write the standard form of the equation of this circle as Standard form

By squaring and simplifying, the equation (x - h)2 + (y - k)2 = r2 can be written in the following general form of the equation of a circle.

Ax2 + Ay2 + Dx + Ey + F = 0, A ≠ 0

To convert such an equation to the standard form

(x - h)2 + (y - k)2 = p

you can use a process called completing the square. If p > 0, the graph of the equation is a circle. If p = 0, the graph is the single point (h, k). If p < 0, the equation has no graph.

Example 2

Completing the Square

Sketch the graph of the circle whose general equation is

4x2 + 4y2 + 20x - 16y + 37 = 0

Solution

To complete the square, first divide by 4 so that the coefficients of x2 and y2 are both 1.

 4x2 + 4y2 + 20x - 16y + 37 = 0 General form x2 + y2 + 5x - 4y = 0 Divide by 4. Group terms. Complete the square by adding and 4 to both sides. = 1 Standard form
Note that you complete the square by adding the square of half the coefficient of x and the square of half the coefficient of y to both sides of the equation. The circle is centered at and its radius is 1, as shown in the figure below.

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