Table of Contents

Last modified on August 3rd, 2023

As we know, a circle is a closed curve drawn from a fixed point called the center. All the points on the curve are having the same distance from the center of the circle. Let’s learn what happens when a circle is graphed on a coordinate plane.

Graphing a circle requires two things, the coordinates of the center point and its radius. There can be two possible situations.

**1) Graphing Circles Centered at the Origin**

Let us put a circle of radius 6 on the graph with the center at (0, 0). In the given circle the radius meets the circumference at (x,y)

Let us work out where the points are exactly on the circle,

Using Pythagoras theorem we get,

x^{2 }+ y^{2} = 6^{2}

Similarly, any point on the circle can be determined using the same rule

x^{2 }+ y^{2} = radius^{2}

**2) Graphing Circles Centered Away from the Origin**

Now let the center be at (h, k), such that it is radius ‘r’ apart from the point (x, y) on the circumference

In order to determine where the point (x, y) is on the circle using the Pythagoras theorem, we get

r^{2} = (x-h)^{2 }+ (y-k)^{2 }

which is the equation of a circle in standard form.

The standard form, also known as the center-radius form of the equation of a circle is given as:

**r ^{2} = (x-h)^{2 }+ (y-k)^{2}**, here (x, y) is any point on the circle, (h, k) is the center, and ‘r’ is the radius

Circles can also be given in the expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.

For example, the equation of a circle with center at (2, 4) and radius 6 is written as,

(x-2)^{2 }+ (y-4)^{2} = 6^{2}

x^{2} – 4x + 4 + y^{2 }– 8y + 16 = 36

x^{2} + y^{2} – 4x – 8y + 20 – 36 = 0

x^{2} + y^{2} – 4x – 8y – 16 = 0

Take a circle having radius ‘r’ and put it in the coordinate plane. Label the circle such that the center is at (h,k) and (x, y) is any point on the circle. The distance between (h, k) and (x, y) is the radius.

Now, we will use the distance formula to calculate the distance between the center (h,k) and the point (x, y) which is the radius ‘r’

Distance formula (d) = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Here,

d = r

(x_{1}, y_{1}) = (h, k)

(x_{2}, y_{2}) = (x, y)

Thus,

r = √(x – h)^{2} + (y – k)

Squaring both sides of the equation, we get

r^{2} = (x – h)^{2} + (y – k)^{2}

=> **(x – h) ^{2} + (y – k)^{2} = r^{2}**

This is the formula for the equation of the circle in standard form. It is also called the center-radius form of the equation of a circle.

We know, the equation of a circle in standard form is:

r = √(x – h)^{2} + (y – k)^{2}

If the center of the above circle is at (0, 0), then the equation can be written as

r = √x^{2} + y^{2}

**r ^{2} = x^{2} + y^{2}**

=> **x ^{2} + y^{2} = r^{2}**

Sometimes, the equation of a circle may not be presented in the neat ‘Standard Form’ as shown above.

As an example, let us put some values to a, b, and rand expand the equation of a circle in standard form.

Let a = 2, b = 3, r= 4, then the equation becomes

(x – 2)^{2} + (y – 3)^{2 }= 4^{2}

x^{2} – 4x + 4 + y^{2} – 6y + 9 = 16

x^{2} + y^{2} – 4x – 6y – 7 = 0

This is also a circle equation but not in the standard form.

Thus, the formula for the equation of the circle in general form is

**x ^{2} + y^{2} + 2gx + 2fy + c = 0**, here (x, y) is any point on the circle, (-g, -f) is the center of the circle, g, f and c are three constants

The radius (r) of such a circle is given as = √g^{2} + f^{2} – c

The basic general equation is this in the form of:

x^{2} + y^{2} + Ax + By + C = 0 (It always has x^{2} + y^{2 }as the first two terms)

Thus the quadratic equation in x and y of the form x^{2} + y^{2} + 2gx + 2fy + c = 0 always represents the equation of a circle.

We know that the equation of the circle having centre at (h, k) and radius ‘r’ is

r^{2} = (x – h)^{2} + (y – k)^{2}

=> r^{2} = x^{2} + y^{2} – 2hx – 2hy + h^{2} + k^{2}

=> x^{2} + y^{2} – 2hx – 2hy + h^{2} + k^{2} – r^{2} = 0

Comparing the above equation with the general equation of the circle x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

h = -g, k = -f and h^{2} + k^{2} – r^{2 }= c

Again, x^{2} + y^{2} + 2gx + 2fy + c = 0

Adding g^{2} + f^{2} – c on both sides, we get

=> (x^{2} + 2gx + g^{2}) + (y^{2} + 2fy + f^{2}) = g^{2} + f^{2} – c

=> (x + g)^{2} + (y + f)^{2} = (√g^{2} + f^{2} – c)^{2}

=> [x – (-g)]^{2} + [(y – (-f)]^{2} = (√g^{2} + f^{2} – c)^{2}

This is of the form (x – h)^{2} + (y – k)^{2} = r^{2} which represents a circle having centre at (- g, -f) and radius √g^{2} + f^{2} – c

Thus, x^{2} + y^{2} – 2hx – 2hy + h^{2} + k^{2} – r^{2} = 0 is the equation of the circle in the general form.

**Conditions**

- If g
^{2}+ f^{2}> c, then the radius of the circle is real - If g
^{2}+ f^{2}= c, then the radius of the circle is zero, which means the circle is a point with coincides with the center. Such a circle is called a point circle. - If g
^{2}+ f^{2}< c, then the radius of the circle is imaginary. Such a circle have a real center and an imaginary radius

Let us solve some problems to find the equation of a circle.

**The equation of a circle is (x – 3) ^{2} + (y – 4)^{2} = 16. Find its center and radius. Graph the circle.**

Solution:

Since the given equation of the circle is in standard form

Its center is at (h, k) = (3, 4)

Radius (r) = √16 = 6 units

The graph is given below

**Write the equation of the circle in standard form having the center at (3, 5) and a radius of 6**

Solution:

As we know, the equation of a circle in standard form is written as

(x-h)^{2 }+ (y-k)^{2} = r^{2}, here (h, k) = (3, 5), r = 6

=> (x-3)^{2 }+ (y-5)^{2} = 6^{2}

**Find the equation of the circle whose center is at origin and radius equal to 9 units?**

Solution:

Given, the center of the circle is (0, 0) and radius = 9 units

As we know, the equation of the circle, with center at origin is

x^{2} + y^{2} = r^{2}

Thus, for the given condition, the equation of a circle is given as

x^{2} + y^{2} = 9^{2}

**The equation of the given circle is x ^{2} + y^{2} − 16x − 12y + 19 = 0. Find the center and radius of the circle.**

Solution:

The given equation is in the form of

x^{2} + y^{2} + 2gx + 2fy + c = 0

Here,

2g = -16

g = -8

2f = -12

f = -6

Thus, the center of the given circle is at (-g, -f) = (8, 6)

Radius = √g^{2} + f^{2} – c, here g = 8, f = 6, c = 19

= √(-8)^{2} + (-6)^{2} – 19

= √64 + 36 – 19

= √81 = 9 units

**Represent the given equation of the circle x ^{2} + y^{2} − 4x − 6y − 4 = 0 in standard form**

Solution:

The given equation of the circle,

x^{2} + y^{2} − 4x − 6y − 4 = 0

Rearranging the above equation we get,

(x^{2} − 4x) + (y^{2} − 6y) − 4 = 0

=> (x^{2} − 4x) + (y^{2} − 6y) = 4

Completing the square for x

=> (x^{2 }– 4x + (-2)^{2}) + (y^{2} − 6y) = 4 + (-2)^{2}

Similarly, completing the square for y

=> (x^{2 }– 4x + (-2)^{2}) + (y^{2} − 6y + (-3)^{2}) = 4 + (-2)^{2 }+ (-3)^{2}

=> (x^{2 }– 4x + 4) + (y^{2} − 6y + 9)

=> (x – 2)^{ 2} + (y – 3)^{ 2}

This is the standard form of the equation of the given circle.

Last modified on August 3rd, 2023