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.

#### Contents

- General Equation of Circle
- Standard Equation of a Circle
- Examples
- Diametric Form

## 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 \(r\).

Locus of P

Now by the distance formula between two points we get

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

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

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

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,

See AlsoEquation of a circle\[\begin{align}\sqrt{2^2 + (-4)^2 - p} &= 5\\4 +16 - p &= 25\\p &= -5. \ _\square\end{align}\]

## 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:

\[\begin{align}x^2-4x+2^2+y^2+2y+1^2&=49+2^2+1^2\\(x^2-4x+4)+(y^2+2y+1)&=49\\(x-2)^2+(y+1)^2&=7^2.\end{align}\]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?

Figure

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

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

Since \(x\) can be equal to \(x_1\) and \(x_2\),

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

## 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

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