10.3 Equations of circles

Circles are fundamental shapes in geometry, defined by all points equidistant from a center. Their equations come directly from the Pythagorean theorem, giving you a precise way to describe and analyze circular shapes on the coordinate plane.

Understanding circle equations lets you identify key features like center coordinates and radius length from an algebraic expression. You'll need this for graphing circles, finding domains and ranges, and solving geometric problems that involve circular shapes.

Equations of Circles

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Derivation of the circle equation

The standard equation of a circle comes from a straightforward application of the distance formula (which itself is built on the Pythagorean theorem). Here's the reasoning:

1. Start with a circle that has center $(h, k)$ and radius $r$.

2. Let $(x, y)$ be any point on the circle. By definition, the distance from $(x, y)$ to the center $(h, k)$ must equal $r$.

3. The horizontal distance between the point and the center is $(x - h)$, and the vertical distance is $(y - k)$. These two distances form the legs of a right triangle, with the radius as the hypotenuse.

4. Applying the Pythagorean theorem gives you:

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

This is the standard form of the equation of a circle. It works for any circle on the coordinate plane:

• $h$ and $k$ are the $x$- and $y$-coordinates of the center
• $r$ is the radius length
• $(x, y)$ represents every point that sits on the circle

Writing circle equations

To write an equation, plug the center coordinates and radius into the standard form $(x - h)^2 + (y - k)^2 = r^2$.

1. Replace $h$ and $k$ with the center's $x$- and $y$-coordinates.

2. Replace $r$ with the radius length (then square it for the right side).

3. Simplify, paying close attention to signs.

Example 1: Center $(-1, 3)$, radius $2$

$(x - (-1))^2 + (y - 3)^2 = 2^2$

which simplifies to $(x + 1)^2 + (y - 3)^2 = 4$

Notice how subtracting a negative flips the sign to a plus. This is the most common place students make errors, so always double-check your signs.

Example 2: Center $(0, 0)$, radius $5$

$(x - 0)^2 + (y - 0)^2 = 5^2$

which simplifies to $x^2 + y^2 = 25$

When the center is the origin, the equation takes this clean, simplified form.

Center and radius from equations

When the equation is already in standard form, you can read the center and radius directly:

• Compare the equation to $(x - h)^2 + (y - k)^2 = r^2$
• The center $(h, k)$ consists of the values being subtracted from $x$ and $y$
• The radius $r$ is the square root of the constant on the right side

Example: $(x - 4)^2 + (y + 2)^2 = 16$

The center is $(4, -2)$ (note that $y + 2$ means $y - (-2)$, so $k = -2$). The radius is $\sqrt{16} = 4$.

When the equation is in general (expanded) form, you need to complete the square to convert it to standard form. This is a critical skill for this unit.

Example: $x^2 + y^2 - 6x + 4y - 12 = 0$

1. Group the $x$-terms and $y$-terms, and move the constant to the right side: $(x^2 - 6x) + (y^2 + 4y) = 12$

2. Complete the square for each group. Take half the linear coefficient, square it, and add it to both sides:

   • For $x$: half of $-6$ is $-3$, and $(-3)^2 = 9$
   • For $y$: half of $4$ is $2$, and $(2)^2 = 4$ $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

3. Factor each perfect square trinomial: $(x - 3)^2 + (y + 2)^2 = 25$

4. Read off the results: center $(3, -2)$, radius $\sqrt{25} = 5$.

A common mistake is forgetting to add the completing-the-square values to both sides of the equation. If you only add them on the left, your radius will be wrong.

Graphing circles on the coordinate plane

To graph a circle from its equation in standard form:

1. Identify the center $(h, k)$ and radius $r$ from the equation.

2. Plot the center point on the coordinate plane.

3. Plot four guide points exactly $r$ units up, down, left, and right from the center:

   • Up: $(h, k + r)$
   • Down: $(h, k - r)$
   • Left: $(h - r, k)$
   • Right: $(h + r, k)$

4. Draw a smooth curve through these four points to form the circle.

The circle is symmetric about both the vertical line $x = h$ and the horizontal line $y = k$ passing through the center.

You can also determine the domain and range directly from the center and radius:

• Domain: $[h - r,\ h + r]$
• Range: $[k - r,\ k + r]$

Example: $(x - 1)^2 + (y + 3)^2 = 4$

• Center: $(1, -3)$, radius: $\sqrt{4} = 2$
• Four guide points: $(1, -1)$, $(1, -5)$, $(-1, -3)$, $(3, -3)$
• Domain: $[-1, 3]$
• Range: $[-5, -1]$