Have you ever found yourself struggling to find the radius of a circle? It can be a frustrating endeavor, but fear not! In this article, we’ll walk through how to solve for the radius of a circle using the equation x^2 + y^2 + 8x – 6y + 21 = 0. So grab a pen and paper, and let’s dive in.

First, let’s review what we know about circles. A circle is a two-dimensional shape that is defined by all points that are equidistant from a fixed point, called the center. The distance from the center to any point on the circle is called the radius. This means that if we can find the center of the circle, we can use the distance formula to calculate the radius.

Now, let’s take a closer look at our equation: x^2 + y^2 + 8x – 6y + 21 = 0. We’ll start by completing the square to put the equation into standard form, which is (x – h)^2 + (y – k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

To complete the square, we first need to group the x-terms and the y-terms together. We can do this by rewriting the equation as follows:

(x^2 + 8x) + (y^2 – 6y) + 21 = 0

Next, we’ll add and subtract terms to both sides of the equation to create perfect squares. To create the perfect square for the x-terms, we need to add (8/2)^2 = 16 to both sides of the equation:

(x^2 + 8x + 16) + (y^2 – 6y) + 21 – 16 = 0

This simplifies to:

(x + 4)^2 + (y^2 – 6y) + 5 = 0

To create the perfect square for the y-terms, we need to add (-6/2)^2 = 9 to both sides of the equation:

(x + 4)^2 + (y^2 – 6y + 9) + 5 – 9 = 0

This simplifies to:

(x + 4)^2 + (y – 3)^2 = 4

Now we have our equation in standard form! We know that (h,k) is the center of the circle and r is the radius. Comparing this to our equation, we see that the center of the circle is (-4,3) and the radius is 2.

To summarize, finding the radius of a circle can be accomplished by completing the square to put the equation into standard form, which is (x – h)^2 + (y – k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. Once the equation is in standard form, we can easily read off the center and radius of the circle.

In conclusion, solving for the radius of a circle can seem daunting at first, but with a little bit of algebraic manipulation, it can become a straightforward task. Completing the square to put the equation into standard form is a powerful tool that allows us to quickly find the center and radius of a circle. So next time you encounter a problem like x^2 + y^2 + 8x – 6y + 21 = 0, you’ll know exactly what to do.