In the figure below, find the equation of the circle.
There are two ways to specify the equation of a circle. We can write the equation in what’s known as center–radius form. Or we can give the expanded form, which is also called the general form of the equation of a circle. Let’s recall the center–radius form. If a circle has its center at the point with coordinates ℎ, 𝑘 and a radius of 𝑟 units, then its equation in center–radius form is given by 𝑥 minus ℎ squared plus 𝑦 minus 𝑘 squared is equal to 𝑟 squared.
To find the equation of our circle then, we just need to determine the coordinates of its center and the length of its radius. The center of our circle is this point here. Looking vertically upwards from this point to the 𝑥-axis, we can see that our center has an 𝑥-coordinate of negative five. And looking horizontally across from this point to the 𝑦-axis, we see that the center has a 𝑦-coordinate of negative four. So the center of our circle has coordinates negative five, negative four. Now, let’s consider the length of its radius.
The radius of a circle is just a straight line joining the center to any point on the circumference. So we can draw in the radius wherever we choose. For example, we can draw in this horizontal radius here. As this line is horizontal, its length will just be the difference between the 𝑥-coordinates of its endpoints, which we can see are negative five and negative 10. So the radius is equal to negative five minus negative 10, which is actually equal to negative five plus 10, which is equal to five. As the horizontal scale on the 𝑥-axis is one square represents one unit, we could also see this by counting the number of squares horizontally from the center to the circumference of the circle.
So, now that we found the center and radius of this circle, we know the values of ℎ, 𝑘, and 𝑟, which we can substitute into the center–radius form of the equation of our circle. We get 𝑥 minus negative five squared plus 𝑦 minus negative four squared is equal to five squared. Now, there is some simplification that we can and should do. In each of our brackets, we have two minus signs. 𝑥 minus negative five is the same as 𝑥 plus five. And 𝑦 minus negative four is the same as 𝑦 plus four. So our brackets simplify to 𝑥 plus five squared and 𝑦 plus four squared. On the right-hand side, we can evaluate five squared, which is equal to 25.
We found then that the equation of the circle shown in the given figure in center–radius form is 𝑥 plus five squared plus 𝑦 plus four squared is equal to 25. If we wanted to give the equation of the circle in its general form, we would need to expand each of the brackets and then simplify the result.