Equation of Circle

Equation of Circle:

Equation of a Circle with given Centre and Radius

We begin by recalling the purpose of coordinate geometry—it expresses geometry algebraically, translating shapes into equations. Consider a circle lying in a plane. Let us introduce a Cartesian coordinate system in this plane.

Assume the circle has its center at point \( C \) and radius \( r \). Let the coordinates of the center \( C \) be \( (\alpha, \beta) \). Let \( P(x, y) \) be any point on the circle.

By the definition of a circle, every point \( P \) on the circle is at a constant distance \( r \) from the center \( C \). That is,

\[ CP = r \]

Using the distance formula between \( C(\alpha, \beta) \) and \( P(x, y) \),

\[ \sqrt{(x - \alpha)^2 + (y - \beta)^2} = r \]

Squaring both sides,

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

This is the required equation of a circle with center \( (\alpha, \beta) \) and radius \( r \). It encodes the entire geometry of the circle into a single algebraic expression.

For example: suppose the center of the circle is \( (2, -1) \) and the radius is \( 3 \). Then applying the general form

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

with \( \alpha = 2 \), \( \beta = -1 \), and \( r = 3 \), we get:

\[ (x - 2)^2 + (y + 1)^2 = 9 \]

This is the required equation of the circle.

When the center of the circle is the origin, that is \( (0, 0) \), and the radius is \( a \), we substitute \( \alpha = 0 \), \( \beta = 0 \), and \( r = a \) into the general form:

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

\[ \implies (x - 0)^2 + (y - 0)^2 = a^2 \]

\[ \implies x^2 + y^2 = a^2 \]

This is the simplest and most symmetric form of the equation of a circle. It represents a circle centered at the origin with radius \( a \).

So, to write the equation of a circle, we need only the center and the radius. The general form

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

contains three unknowns: \( \alpha \), \( \beta \), and \( r \). These represent the coordinates of the center and the radius. Hence, any problem that gives enough information to determine these three values allows us to write the complete equation of the circle.

Let us now work through some examples where the given geometric information helps us deduce \( \alpha \), \( \beta \), and \( r \), and then use these to write the equation of the circle.

Example

A circle passes through the origin and has center at \( (4, 3) \). Find its equation.

Solution:

The center is given: \( \alpha = 4, \beta = 3 \).

Since the circle passes through the origin \( (0, 0) \), the distance from the center to the origin gives the radius:

\[ r = \sqrt{(0 - 4)^2 + (0 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Now substitute into the general form:

\[ (x - 4)^2 + (y - 3)^2 = 25 \]

Example

A circle has diameter joining the points \( (1, 4) \) and \( (5, 2) \). Find its equation.

Solution:

The center is the midpoint of the diameter:

\[ \alpha = \frac{1 + 5}{2} = 3, \quad \beta = \frac{4 + 2}{2} = 3 \]

The radius is half the length of the diameter:

\[ r = \frac{1}{2} \cdot \sqrt{(5 - 1)^2 + (2 - 4)^2} = \frac{1}{2} \cdot \sqrt{16 + 4} = \frac{1}{2} \cdot \sqrt{20} = \sqrt{5} \]

Hence, the equation is:

\[ (x - 3)^2 + (y - 3)^2 = 5 \]

Example

Find the equation of a circle touching the \( x \)-axis at \( (3, 0) \) and having radius \( 5 \).

Solution:

Plotting the given information on a coordinate plane helps us visualize the geometry. The circle touches the \( x \)-axis at the point \( (3, 0) \). Since a circle touches a line at exactly one point, the radius at the point of contact must be perpendicular to the tangent, which in this case is the \( x \)-axis. Therefore, the center of the circle lies vertically above the point \( (3, 0) \), i.e., it must be on the line \( x = 3 \).

Given that the radius is \( 5 \), the vertical distance from the point \( (3, 0) \) to the center must be \( 5 \). Hence, the center is at:

\[ (\alpha, \beta) = (3, 5) \]

Using the standard equation of a circle:

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

we substitute \( \alpha = 3 \), \( \beta = 5 \), and \( r = 5 \):

\[ (x - 3)^2 + (y - 5)^2 = 25 \]

This is the required equation of the circle.

Example

Find the equation of a circle of radius \( 5 \) that touches the \( x \)-axis and whose center lies on the line \( y = x - 1 \), in the first quadrant.

Solution:
Plotting the graph, we observe that the center of the circle lies somewhere on the line \( y = x - 1 \), and the circle touches the \( x \)-axis. Since it touches the \( x \)-axis, the vertical distance from the center to the axis must equal the radius of the circle.

Let the center be \( (\alpha, \alpha - 1) \). This is valid since the center lies on the line \( y = x - 1 \); if the abscissa is \( \alpha \), the ordinate must be \( \alpha - 1 \).

Now, because the circle touches the \( x \)-axis, the distance from the center to the \( x \)-axis is equal to the radius:

\[ \alpha - 1 = 5 \implies \alpha = 6 \]

Thus, the center is at \( (6, 5) \), and the radius is \( 5 \). Therefore, the required equation of the circle is

\[ (x - 6)^2 + (y - 5)^2 = 25 \]

Example

Find the equation of the circle with minimum radius that touches the \( y \)-axis and passes through the point \( (3, 4) \).

Solution:

There are infinitely many circles that pass through \( (3, 4) \) and touch the \( y \)-axis. But among them, the one with minimum radius is the one where the point \( (3, 4) \) and the point of contact on the \( y \)-axis are diametrically opposite.

From the geometry, such a circle will have the diameter joining \( (3, 4) \) and the point \( P \) on the \( y \)-axis. Since the point lies on the \( y \)-axis and must have the same ordinate to be diametrically opposite (to minimize radius), \( P \) must be \( (0, 4) \).

Then the diameter is the segment joining \( (3, 4) \) and \( (0, 4) \), and hence:

\[ \text{Radius} = \frac{1}{2} \times \text{length of diameter} = \frac{1}{2} \times 3 = \frac{3}{2} \]

The center is the midpoint of these two points:

\[ \left( \frac{3 + 0}{2}, \frac{4 + 4}{2} \right) = \left( \frac{3}{2}, 4 \right) \]

Now using the standard form of the equation of a circle with center \( (\alpha, \beta) \) and radius \( r \):

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

Substitute \( \alpha = \frac{3}{2} \), \( \beta = 4 \), \( r = \frac{3}{2} \):

\[ \left( x - \frac{3}{2} \right)^2 + (y - 4)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \]

This is the required equation of the circle with the minimum possible radius.

Equation of a Circle When Two Diametrically Opposite Points Are Known

Let \( A(x_1, y_1) \) and \( B(x_2, y_2) \) be the endpoints of a diameter of a circle.
A naïve approach to find the equation is as follows:

The radius is half the distance between \( A \) and \( B \):

\[ r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

The center is the midpoint of \( AB \):

\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Substituting this into the standard form gives:

\[ \left( x - \frac{x_1 + x_2}{2} \right)^2 + \left( y - \frac{y_1 + y_2}{2} \right)^2 = \left( \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \right)^2 \]

\[ \implies \left( x - \frac{x_1 + x_2}{2} \right)^2 + \left( y - \frac{y_1 + y_2}{2} \right)^2 = \frac{1}{4} \left[ (x_2 - x_1)^2 + (y_2 - y_1)^2 \right] \]

This is correct, but there exists a more elegant and geometrically insightful method.

Let \( P(x, y) \) be any arbitrary point on the circle. Since \( AB \) is a diameter, the angle \( \angle APB \) is a right angle. Thus,

\[ \text{AP} \perp \text{PB} \]

Using slopes:

\[ \text{Slope of } AP = \frac{y - y_1}{x - x_1}, \quad \text{Slope of } PB = \frac{y - y_2}{x - x_2} \]

As the angle is \( 90^\circ \), the product of these slopes is \( -1 \):

\[ \frac{y - y_1}{x - x_1} \cdot \frac{y - y_2}{x - x_2} = -1 \]

\[ \implies (y - y_1)(y - y_2) + (x - x_1)(x - x_2) = 0 \]

This is the required equation of the circle when two diametrically opposite points are known.

Example

Find the equation of the circle whose diameter has endpoints \( A(1, 2) \) and \( B(5, 4) \).

Using the derived form:

\[ (x - 1)(x - 5) + (y - 2)(y - 4) = 0 \]

\[ \implies x^2 - 6x + 5 + y^2 - 6y + 8 = 0 \]

\[ \implies x^2 + y^2 - 6x - 6y + 13 = 0 \]

This is the equation of the circle with diameter \( AB \).

General Equation of a Circle

Consider a circle with center \( (\alpha, \beta) \) and radius \( r \). Then its equation is:

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

Expanding this:

\[ x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 = r^2 \]

\[ \implies x^2 + y^2 - 2\alpha x - 2\beta y + (\alpha^2 + \beta^2 - r^2) = 0 \]

This form involves: - Quadratic terms \( x^2 + y^2 \), - Linear terms \( x \) and \( y \), - A constant term.

So we claim that any equation of a circle can be written in the form:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

The choice of coefficients as \( 2g \) and \( 2f \) rather than \( g \) and \( f \) is intentional, as it simplifies the formulas that follow—especially for the center and radius.

But here is an important caveat:

Not every equation of the form

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

represents a circle.

Let us determine the condition under which it does.

We rewrite the equation by completing the square:

\[ (x^2 + 2gx) + (y^2 + 2fy) = -c \]

Add \( g^2 + f^2 \) to both sides to complete the squares on the left:

\[ (x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) = -c + g^2 + f^2 \]

\[ \implies (x + g)^2 + (y + f)^2 = g^2 + f^2 - c \]

Comparing with the standard form:

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

we conclude that this equation represents a circle with

• Center: \( (-g, -f) \)
• Radius: \( r = \sqrt{g^2 + f^2 - c} \)

But the square root is defined only if \( g^2 + f^2 - c \geq 0 \).

Thus, the equation

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

represents a circle if and only if:

\[ g^2 + f^2 - c \geq 0 \]

and equality holds if and only if the circle reduces to a point. If \( g^2 + f^2 - c < 0 \), the equation has no real solution and represents no real circle.

The case when

\[ g^2 + f^2 - c = 0 \]

is rather peculiar. It means the radius of the circle is zero, so the equation becomes:

\[ (x + g)^2 + (y + f)^2 = 0 \]

Now, since the sum of two squares is zero only when both squares are zero, this equation is satisfied by exactly one point:

\[ x = -g, \quad y = -f \]

Thus, the only solution to the equation is the single point \( (-g, -f) \), which is also the center of the circle. This degenerate case is called a point circle.

It is a limiting case of a circle where the entire geometry collapses to a single point, but algebraically it still satisfies the form of the general equation:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

with \( g^2 + f^2 - c = 0 \).

Reading the equation of a Circle

We can now read directly the center and radius of a circle from its equation by mentally comparing it to the general form:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

For example, consider the equation:

\[ x^2 + y^2 - 4x + 6y - 1 = 0 \]

We observe and identify: [ 2g = -4 \implies g = -2, \quad 2f = 6 \implies f = 3, \quad c = -1 ]

Then the center is:

\[ (-g, -f) = (2, -3) \]

and the radius is:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + 3^2 - (-1)} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

Thus, without any need to complete the square, we immediately conclude that the circle has center \( (2, -3) \) and radius \( \sqrt{14} \).

Another Example

Consider the equation:

\[ 3x^2 + 3y^2 - 2x - 4y - 1 = 0 \]

This is not in standard form. To bring it to the standard form of a circle, we first observe that the coefficients of \( x^2 \) and \( y^2 \) must be 1. So we divide the entire equation by 3:

\[ x^2 + y^2 - \frac{2}{3}x - \frac{4}{3}y - \frac{1}{3} = 0 \]

Now, comparing with the standard general form:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

we extract:

\[ 2g = -\frac{2}{3} \implies g = -\frac{1}{3}, \quad 2f = -\frac{4}{3} \implies f = -\frac{2}{3}, \quad c = -\frac{1}{3} \]

Then the center is:

\[ (-g, -f) = \left( \frac{1}{3}, \frac{2}{3} \right) \]

and the radius is:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{1}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 - \left(-\frac{1}{3}\right)} = \sqrt{\frac{1}{9} + \frac{4}{9} + \frac{1}{3}} = \sqrt{ \frac{5}{9} + \frac{3}{9} } = \sqrt{ \frac{8}{9} } = \frac{2\sqrt{2}}{3} \]

Thus, the circle has center \( \left( \frac{1}{3}, \frac{2}{3} \right) \) and radius \( \frac{2\sqrt{2}}{3} \).

Example

Find the equation of the circle passing through the points \( (1, 1) \), \( (0, 1) \), and \( (3, 0) \).

Solution:

In such problems, the general form of the circle is extremely helpful. Let us assume the equation of the circle to be:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

There are three unknowns in this equation: \( g, f, c \). To determine them, we need three equations. These can be obtained directly by substituting the three given points, which must satisfy the circle’s equation.

Using \( (1, 1) \):

\[ 1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 1 + 1 + 2g + 2f + c = 0 \implies 2g + 2f + c = -2 \tag{1} \]

Using \( (0, 1) \):

\[ 0^2 + 1^2 + 2g(0) + 2f(1) + c = 0 \implies 1 + 2f + c = 0 \implies 2f + c = -1 \tag{2} \]

Using \( (3, 0) \):

\[ 3^2 + 0^2 + 2g(3) + 2f(0) + c = 0 \implies 9 + 6g + c = 0 \implies 6g + c = -9 \tag{3} \]

Now solve the system:

Subtract (2) from (1):

\[ (2g + 2f + c) - (2f + c) = -2 - (-1) \implies 2g = -1 \implies g = -\frac{1}{2} \]

Substitute \( g = -\frac{1}{2} \) into (3):

\[ 6 \cdot \left(-\frac{1}{2}\right) + c = -9 \implies -3 + c = -9 \implies c = -6 \]

Substitute \( c = -6 \) into (2):

\[ 2f - 6 = -1 \implies 2f = 5 \implies f = \frac{5}{2} \]

Thus, the required equation is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

\[ \implies x^2 + y^2 - x + 5y - 6 = 0 \]

This is the equation of the circle passing through the three given points.

Parametric Form of a Circle

Consider a circle with center at \( (\alpha, \beta) \) and radius \( r \). Its equation is:

\[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]

We want to extract a parametric form of this equation—that is, a representation of all points on the circle in terms of a single real parameter. To do this, divide both sides of the equation by \( r^2 \):

\[ \left( \frac{x - \alpha}{r} \right)^2 + \left( \frac{y - \beta}{r} \right)^2 = 1 \]

This resembles the identity:

\[ \cos^2 \theta + \sin^2 \theta = 1 \]

Thus, define a real parameter \( \theta \) such that

\[ \frac{x - \alpha}{r} = \cos \theta, \qquad \frac{y - \beta}{r} = \sin \theta \]

Solving for \( x \) and \( y \), we obtain:

\[ x = \alpha + r \cos \theta, \qquad y = \beta + r \sin \theta \]

This is the required parametric form of the circle. Every point \( (x, y) \) lying on the circle can be expressed in this way for some \( \theta \in \mathbb{R} \).

To interpret this parametrization geometrically, draw a horizontal line through the center of the circle, and let \( P \) be a point on the circle. Drop a perpendicular from \( P \) to this horizontal line, with foot at \( N \). Let \( C \) denote the center of the circle.

The angle \( \theta \) is the angle \( \angle PCN \) that the radius \( \overrightarrow{CP} \) makes with the positive direction of the horizontal line (i.e., the x-axis). Then:

\[ CN = r \cos \theta, \qquad PN = r \sin \theta \]

But in terms of coordinates, \( CN = x - \alpha \), and \( PN = y - \beta \). Hence:

\[ x - \alpha = r \cos \theta, \qquad y - \beta = r \sin \theta \]

which confirms:

\[ (x, y) = (\alpha + r \cos \theta,\ \beta + r \sin \theta) \]

Therefore, every point on the circle can be expressed in this form for some real \( \theta \).

For example, consider the circle with center at the origin and radius \( a \). Its equation is:

\[ x^2 + y^2 = a^2 \]

Here, \( \alpha = 0 \), \( \beta = 0 \), and \( r = a \). Substituting into the parametric form, we obtain:

\[ x = a \cos \theta, \qquad y = a \sin \theta \]

Thus, the point \( (x, y) \) lies on the circle if and only if it can be written as \( (a \cos \theta,\ a \sin \theta) \) for some \( \theta \in \mathbb{R} \).
This parametrization traces the circle as \( \theta \) varies over \( [0, 2\pi] \), and repeats periodically as \( \theta \) increases further.

In this form, the variable \( \theta \) has a clear geometric interpretation: it is the angle between the positive x-axis and the radius vector from the origin to the point \( (x, y) \), measured in the counter-clockwise direction.

This parametric form is particularly useful in situations where we wish to introduce trigonometry into algebraic problems, especially when it makes conceptual and computational sense to do so. A common setting is in maximization and minimization problems involving points on a circle.

In such cases, trigonometric expressions are often easier to differentiate or manipulate than Cartesian expressions involving both \( x \) and \( y \). Moreover, the parametric form transforms a problem with two variables—the abscissa and the ordinate—into a problem with only one variable, namely \( \theta \). This reduction in the number of unknowns simplifies both the reasoning and the calculation.

The advantage becomes clearer when we examine the following examples.

Example

A point \( (x, y) \) satisfies the equation \( (x - 1)^2 + (y - 2)^2 = 9 \).
Find the maximum and minimum value of the expression \( 3x + 4y \).

Solution:

We recognize that the given equation represents a circle with
center \( (1, 2) \) and radius \( 3 \).

To evaluate the maximum and minimum values of a linear expression over this circle, we introduce the parametric form of a point on the circle:

\[ x = 1 + 3 \cos \theta, \qquad y = 2 + 3 \sin \theta \quad \text{for } \theta \in \mathbb{R} \]

Substituting into the expression \( 3x + 4y \), we get:

\[ 3x + 4y = 3(1 + 3 \cos \theta) + 4(2 + 3 \sin \theta) \]

\[ = 3 + 9 \cos \theta + 8 + 12 \sin \theta = 11 + 9 \cos \theta + 12 \sin \theta \]

Now, recall the standard inequality:

\[ - \sqrt{a^2 + b^2} \leq a \cos \theta + b \sin \theta \leq \sqrt{a^2 + b^2} \]

Applying this with \( a = 9 \), \( b = 12 \), we obtain:

\[ - \sqrt{9^2 + 12^2} \leq 9 \cos \theta + 12 \sin \theta \leq \sqrt{9^2 + 12^2} \]

\[ \implies -15 \leq 9 \cos \theta + 12 \sin \theta \leq 15 \]

Therefore,

\[ 11 - 15 \leq 3x + 4y \leq 11 + 15 \implies -4 \leq 3x + 4y \leq 26 \]

Hence, the maximum value of \( 3x + 4y \) is \( \boxed{26} \), and the minimum value is \( \boxed{-4} \).