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.

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.

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

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

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 $x^2 + y^2 - 6x - 8y + 10$?

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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 $p$.

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

Using the above equation,

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

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 $(x-3)^2+(y-5)^2=49$.

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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?

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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 $x^2-4x+y^2+2y=44$?

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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 $x^2+y^2=25$?

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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 origin and its radius is $5$! $_\square$

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

Figure

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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,

$$k = 3.\ _\square$$

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 $(-4, 3).$

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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

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