Circle Equations
Move the point \((x, y)\) around. What do you notice? What to you wonder? The right triangle has legs that are \(x\) and \(y\) units long, and the hypotenuse is \(r\) units long. We can use the Pythagorean Theorem: \(leg^2 + leg^2 = hypotenuse^2\) Substituting the lengths into the equation: \(x^2 + y^2 = r^2\) We could use the distance formula instead with the points \((x, y)\) and the origin: \(distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Substituting the coordinates into the equation: \(r = \sqrt{(x - 0)^2 + (y - 0)^2}\) If we square both sides of the equation: \(r^2 = (x - 0)^2 + (y - 0)^2 = x^2 + y^2\) Therefore circles with their centers at the origin, \((0, 0)\), have the equation \(x^2 + y^2 = r^2\) This means that any point \((x, y)\) on the circle is \(r\) units away from the center. |
|
What if the center of the circle is not at \((0, 0)\)? Given a circle with center at \((h, k)\), our new right triangle looks like the diagram at the right.
Using the Pythagorean Theorem: \((x - h)^2 + (y - k)^2 = r^2\) We could use the distance formula instead: \(r = \sqrt{(x - h)^2 + (y - k)^2}\) If we square both sides of the equation: \(r^2 = (x - h)^2 + (y - k)^2\) Both methods produce the standard form of the equation of a circle. |
Standard Form of the Equation of a Circle with center at \((h, k)\) and radius \(r\)
\((x - h)^2 + (y - k)^2 = r^2\)
\((x - h)^2 + (y - k)^2 = r^2\)
Let's try some examples.
Example 1:
What are the center and the radius for the circle \((x - 7)^2 + (y - 1)^2 = 25\)
Solution:
Center at \((h, k) = (7, 1)\) and a radius of \(\sqrt{25} = 5\)
Equations of circles can be written in other forms. It is helpful to determine the center and the radius if a circle is written in standard form. To rewrite an equation in standard form, completing the square may need to be used. Let's look at an example.
Example 2:
Write the equation in standard form and determine the radius and center of the circle: \(x^2 + y^2 + 2x + 8y + 10 = 0\)
Solution
\(\begin{align*}
x^2 + 2x + y^2 + 8y &= -10 & \text{subract \(10\) from both sides}\\
x^2 + 2x + \;\;?\;\; + y^2 + 8y + \;\;?\;\; &= -10 + \;\;?\;\; + \;\; ?\;\; & \text{create space for the values to complete the square}\\
x^2 + 2x + 1 + y^2 + 8y + 16 &= -10 + 1 + 16 & \text{add $1$ and $16$ to create perfect square trinomials}\\
(x + 1)^2 + (y + 4)^2 &= 7\\
\end{align*}\)
Center: \((-1, -4)\) radius = \(\sqrt{7}\)
Example 3:
Write the equation in standard form and determine the radius and center of the circle: \(x^2 + 4x + y^2 - 10y = 7\)
Solution:
Let's watch a video for the solution.
Example 1:
What are the center and the radius for the circle \((x - 7)^2 + (y - 1)^2 = 25\)
Solution:
Center at \((h, k) = (7, 1)\) and a radius of \(\sqrt{25} = 5\)
Equations of circles can be written in other forms. It is helpful to determine the center and the radius if a circle is written in standard form. To rewrite an equation in standard form, completing the square may need to be used. Let's look at an example.
Example 2:
Write the equation in standard form and determine the radius and center of the circle: \(x^2 + y^2 + 2x + 8y + 10 = 0\)
Solution
\(\begin{align*}
x^2 + 2x + y^2 + 8y &= -10 & \text{subract \(10\) from both sides}\\
x^2 + 2x + \;\;?\;\; + y^2 + 8y + \;\;?\;\; &= -10 + \;\;?\;\; + \;\; ?\;\; & \text{create space for the values to complete the square}\\
x^2 + 2x + 1 + y^2 + 8y + 16 &= -10 + 1 + 16 & \text{add $1$ and $16$ to create perfect square trinomials}\\
(x + 1)^2 + (y + 4)^2 &= 7\\
\end{align*}\)
Center: \((-1, -4)\) radius = \(\sqrt{7}\)
Example 3:
Write the equation in standard form and determine the radius and center of the circle: \(x^2 + 4x + y^2 - 10y = 7\)
Solution:
Let's watch a video for the solution.
For more examples on general completing the square, click this link to completing the square from the Algebra 1 Digi (scroll down):
Example 4:
Write the equation of the circle with center \((2, -3)\) radius = \(9\)
Solution:
\((x - 2)^2 + (y + 3)^2 = 81\)
Example 5:
Write an equation of the circle with center \((3, 5)\) point \((-3, -3)\)
radius = \(\sqrt{(3 + 3)^2 + (5 + 3)^2}\)
radius = \(\sqrt{36 + 64}\) = \(\sqrt{100} = 10\)
\((x - 3)^2 + (y - 5)^2 = 100\)
Example 6:
Write an equation of the circle given two endpoints of a diameter \((-6, 6)\) and \((0, 6)\)
Solution:
We need to find the midpoint of the two points. This will give us the center of the circle.
\(\left(\Large\frac{-6 + 0}{2}, \frac{6 + 6}{2}\right) = (-3, 6)\)
Then we need to find the radius which is the distance between the center and one of the points.
\(r = \sqrt{(-6 + 3)^2 + (6 - 6)^2} = \sqrt{(-3)^2 + (0)^2} = \sqrt{9} = 3\)
The equation using both the center and radius is:
\((x + 3)^2 + (y - 6)^2 = 9\)
Graphing Circles:
Example 4:
Write the equation of the circle with center \((2, -3)\) radius = \(9\)
Solution:
\((x - 2)^2 + (y + 3)^2 = 81\)
Example 5:
Write an equation of the circle with center \((3, 5)\) point \((-3, -3)\)
radius = \(\sqrt{(3 + 3)^2 + (5 + 3)^2}\)
radius = \(\sqrt{36 + 64}\) = \(\sqrt{100} = 10\)
\((x - 3)^2 + (y - 5)^2 = 100\)
Example 6:
Write an equation of the circle given two endpoints of a diameter \((-6, 6)\) and \((0, 6)\)
Solution:
We need to find the midpoint of the two points. This will give us the center of the circle.
\(\left(\Large\frac{-6 + 0}{2}, \frac{6 + 6}{2}\right) = (-3, 6)\)
Then we need to find the radius which is the distance between the center and one of the points.
\(r = \sqrt{(-6 + 3)^2 + (6 - 6)^2} = \sqrt{(-3)^2 + (0)^2} = \sqrt{9} = 3\)
The equation using both the center and radius is:
\((x + 3)^2 + (y - 6)^2 = 9\)
Graphing Circles:
We can also graph circles and find equations of circles from the graphs.
In the example above the center of the circle of the purple circle is \((5, -4)\) and the radius is \(4\) units. Therefore the equation of the circle is \((x - 5)^2 + (y + 4)^2 = 16\). Try finding the equations of the blue & red circles.
Quick Check
1) Write the equation of a circle given center \((2, -7)\) radius = \(11\)
2) Write the equation of a circle given a center \((1, -6)\) point \((4, -7)\)
Quick Check Solutions
1) Write the equation of a circle given center \((2, -7)\) radius = \(11\)
2) Write the equation of a circle given a center \((1, -6)\) point \((4, -7)\)
Quick Check Solutions