# Equation of a Circle

A circle is the set of all points in the plane which maintains a fixed finite distance $r$ from a fixed point $O=(a,b)$. Here $O$ is called the center, and $r$ is called the radius of that circle. The use of the equation of a circle is prevalent throughout coordinate geometry problems.

## General Equation of Circle

The equation of any conic can be expressed as

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0.$

However, the condition for the equation to represent a circle is $a = b$ and $h = 0$. Then the general equation of the circle becomes

$x^2 + y^2 + 2gx + 2fy + c = 0.$

Unfortunately, it can be difficult to decipher any meaningful properties about a given circle from its general equation, so completing the square allows quick conversion to the standard form, which contains values for the center and radius of the circle.

## Standard Equation of a Circle

The standard equation for a circle contains pertinent information about the circle's center and radius and is therefore much easier to read at a glance.

The standard equation of a circle with center at $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2$= $r^2$.

If we have a point $O=(a,b)$ in the plane and a radius $r$, then we can construct a unique circle.

We find the locus of a point which moves in such a way that its distance $r$ from another point ($a,b$) is always constant. Now if $P=(h,k)$ is any point on the unique circle with center $O$ and radius $r$, the distance from $O=(a,b)$ to $P=(h,k)$ must be $r$.

Now by the distance formula between two points we get

$\begin{aligned} \sqrt { (h-a)^{ 2 }+(k-b)^{ 2 } } &=r\\ (h-a)^{ 2 }+(k-b)^{ 2 }&=r^{ 2 }. \end{aligned}$

Replacing $h$ by $x$ and $k$ by $y,$ we get

$(x-a)^2+(y-b)^2=r^2.\ _\square$

This is the standard equation of a circle, with radius $r$ and center at $(a,b)$.

The general form can be converted into the standard form by completing the square. First, we combine like terms

$\begin{aligned} x^2 + 2gx + y^2 + 2fy + c &= 0\\ (x+g)^2 + (y+f)^2 - g^2 - f^2 + c &= 0\\ (x+g)^2 + (y+f)^2 &= g^2 + f^2 - c. \end{aligned}$

Comparing this with the standard form, we come to know that

- center $= (-g, -f)$
- radius $= \sqrt{g^2 + f^2 -c}.$

if $g^2 + f^2 -c> 0,$ it is a real circle.

if $g^2 + f^2 -c =0,$ it is a point circle.

if $g^2 + f^2 -c <0,$ it is an imaginary circle with no locus.

However, make sure the coefficients of $x^2$ and $y^2$ are 1 before applying these formulae.

## What are the coordinates of the center of the circle $x^2 + y^2 - 6x - 8y + 10$?

The coordinates of the center of the circle are $(-g, -f)$. Here $g = -\frac{6}{2} = - 3$ and $f = - \frac{8}{2}= -4$, so center $= (3, 4).$ $_\square$

## Given that the circle $x^2 + y^2 + 4x - 8y + p = 0$ has radius 5, find the possible value of $p$.

In this circle, $g = 2 , f = -4 , c = p$, and radius = $5$.

Using the above equation,

$\begin{aligned} \sqrt{2^2 + (-4)^2 - p} &= 5\\ 4 +16 - p &= 25\\ p &= -5. \ _\square \end{aligned}$

## Examples

While solving problems, we try to make the left-hand side of the form $(x-a)^2+(y-b)^2$ by using completing the square method.

## Draw the graph of the equation $(x-3)^2+(y-5)^2=49$.

Notice that the right side is $7^2$. Comparing to the standard equation of a circle, we easily see that the graph is a circle with radius $7$ and center at $(3,5)$. Now we can easily draw the graph using compass. $_\square$

## What does the graph of the equation $x^2+y^2-2x-14y+34=0$ look like?

Notice that we can rewrite the equation as

$\left(x^2-2x+1\right)+\left(y^2-14y+49\right)=16.$

Completing the squares, this becomes

$(x-1)^2+(y-7)^2=4^2.$

So the graph is a circle with radius $4$, centered at $(1,7)$. $_\square$

## What are the radius and center of the circle whose equation is $x^2-4x+y^2+2y=44$?

We can apply completing the square method to the left-hand side:

$\begin{aligned} x^2-4x+2^2+y^2+2y+1^2&=49+2^2+1^2\\ (x^2-4x+4)+(y^2+2y+1)&=49\\ (x-2)^2+(y+1)^2&=7^2. \end{aligned}$ Comparing with the standard equation, we can see that $a=2$ and $b=-1.$ Therefore, the center of the circle is $(2,-1)$ and its radius is $7.$ $_\square$

## What are the radius and center of the circle whose equation is $x^2+y^2=25$?

We can rewrite the given equation as

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

Comparing with the standard equation, we can see that $a=b=0.$ Therefore the center of the circle is the

originand its radius is $5$! $_\square$

What is the value of $k$ in the figure below?

Since it is a circle and is touching both the $x$-axis and $y$-axis, its distance from both the axes must be the same. Since it is $3$ units away from the $x$-axis, it must be $3$-units away from the $y$-axis. Therefore,

$k = 3.\ _\square$

## Diametric Form

Another way of expressing the equation of a circle is the diametric form.

Suppose there are two points on a circle $(x_1, y_1)$ and $(x_2, y_2)$, such that they lie on the opposite ends of the same diameter, then the equation of the circle can be written as

$(x-x_1)(x-x_2) + (y-y_1)(y - y_2) = 0.$

Suppose 2 points on the circle $A= (x_1, y_1)$ and $B= (x_2, y_2)$ are diametrically opposite, then for any point $C= (x, y)$ on the circle, $\triangle ABC$ will be a right triangle with right angle at $C$. This implies

$\begin{aligned} AC &\perp BC\\ (m_{AC}) \cdot (m_{BC}) &= -1\\ \left(\dfrac{y - y_1}{x - x_1}\right) \cdot \left( \dfrac{y - y_2}{x - x_2}\right)&= -1. \end{aligned}$

Since $x$ can be equal to $x_1$ and $x_2$,

$\begin{aligned} (y-y_1)(y - y_2 ) &= - (x - x_1) (x- x_2)\\ (x-x_1)(x-x_2) + (y-y_1)(y - y_2) &= 0. \ _\square \end{aligned}$

## Find the equation of the smallest possible circle that passes through the points $(2,6)$ and $(-4, 3).$

The circle would be the smallest if the two points were to be the endpoints of a diameter of the circle.

We can use the diametric form to get

$\begin{aligned} (x - 2)(x - (-4)) + (y - 6)(y - 3) &=0\\ (x - 2)(x +4) + (y - 6)(y - 3) &=0. \ _\square \end{aligned}$

**Cite as:**Equation of a Circle.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/conics-circle-standard-equation/