Welcome to the second lesson on circles. This lesson will talk about various forms of equations to a circle in the Cartesian plane. Let’s begin!

## Equation of a Circle with centre (x_{1}, y_{1}) and radius r

Let P(x, y) be any point on the circle. Then, following the definition of the circle, we have CP = r

Using distance formula, we get the required equation as \( \sqrt{(x-x_1)^2+(y-y_1)^2}=r\), or **(x – x _{1})^{2} + (y – y_{1})^{2} = **r

^{2}

## General Equation of a Circle

If you expand the equation obtained just now, you’ll get something like this: x^{2} + y^{2} – 2x_{1}x – 2y_{1}y + x_{1}^{2} + y_{1}^{2} – r^{2} = 0

The equation has two terms of second degree: x^{2} and y^{2}, which have the same coefficient (1 in this case); two terms of first degree: – 2x_{1}x and – 2y_{1}y; and one constant term.

Now consider this equation: x^{2} + y^{2} + 2gx + 2fy + c = 0, which looks somewhat like the one above (i.e. the number and types of second degree, first degree and constant terms)

Will this represent a circle? (because it looks like the equation of a circle?). We did something like this when we were talking about the general equation of a straight line.

Turns out it will. We just have to change its form to the one like we just derived.

I’ve added and subtracted a few terms to the equation, and rearranged the terms, so that it looks like this: x^{2} + 2gx + g^{2} + y^{2} + 2fy + f^{2} = g^{2} + f^{2} – c

The equation can now be written as (x + g)^{2} + (y + f)^{2} = (\(\sqrt{g^2+f^2-c}\))^{2} which will look exactly like (x – x_{1})^{2} + (y – y_{1})^{2} = r^{2}, on substituting g = – x_{1}, f = – y_{1} and g^{2} + f^{2} – c = r^{2}

Now since (x – x_{1})^{2} + (y – y_{1})^{2} = r^{2} represents a circle whose center is (x_{1}, y_{1}) and radius r, therefore, x^{2} + y^{2} + 2gx + 2fy + c = 0 will also represent a circle whose center is (–g, –f) and radius equals \(\sqrt{g^2+f^2-c}\) (using the substitutions we made)

The only thing we need to ensure that g^{2} + f^{2} – c must be positive for the radius to be defined. In case g^{2} + f^{2} – c is zero, the radius of the circle also becomes zero. In this case, the circle is called a **point circle** (something which you needn’t be worried about right now)

## Lesson Summary

- Equation of the circle whose center is (x
_{1}, y_{1}) and radius is r is given by**(x – x**_{1})^{2}+ (y – y_{1})^{2}= r^{2} - The general equation of a circle is
**x**, whose center is^{2}+ y^{2}+ 2gx + 2fy + c = 0**(– g, – f)**and radius is \(\sqrt{g^2+f^2-c}\)

The next lesson will cover a few examples related to equations of the circle.