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.
Circle on a Graph
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)
Graphing Circles Centered at the Origin
Let us work out where the points are exactly on the circle,
Using Pythagoras theorem we get,
x2 + y2 = 62
Similarly, any point on the circle can be determined using the same rule
x2 + y2 = radius2
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
Graphing Circles Centered Away from the Origin
In order to determine where the point (x, y) is on the circle using the Pythagoras theorem, we get
r2 = (x-h)2 + (y-k)2
which is the equation of a circle in standard form.
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:
r2 = (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
Equation of a Circle
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 = 62
x2 – 4x + 4 + y2 – 8y + 16 = 36
x2 + y2 – 4x – 8y + 20 – 36 = 0
x2 + y2 – 4x – 8y – 16 = 0
How to Derive the Equation of a Circle in Standard Form
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) = √(x2 – x1)2 + (y2 – y1)2
Here,
d = r
(x1, y1) = (h, k)
(x2, y2) = (x, y)
Thus,
r = √(x – h)2 + (y – k)
Squaring both sides of the equation, we get
r2 = (x – h)2 + (y – k)2
=> (x – h)2 + (y – k)2 = r2
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.
Equation of a Circle in Standard Form (Centered at the Origin)
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 = √x2 + y2
r2 = x2 + y2
=> x2 + y2 = r2
Equation of a Circle in General Form
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 = 42
x2 – 4x + 4 + y2 – 6y + 9 = 16
x2 + y2 – 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
x2 + y2 + 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 = √g2 + f2 – c
The basic general equation is this in the form of:
x2 + y2 + Ax + By + C = 0 (It always has x2 + y2 as the first two terms)
Thus the quadratic equation in x and y of the form x2 + y2 + 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
r2 = (x – h)2 + (y – k)2
=> r2 = x2 + y2 – 2hx – 2hy + h2 + k2
=> x2 + y2 – 2hx – 2hy + h2 + k2 – r2 = 0
Comparing the above equation with the general equation of the circle x2 + y2 + 2gx + 2fy + c = 0, we get
This is of the form (x – h)2 + (y – k)2 = r2 which represents a circle having centre at (- g, -f) and radius √g2 + f2 – c
Thus, x2 + y2 – 2hx – 2hy + h2 + k2 – r2 = 0 is the equation of the circle in the general form.
Conditions
If g2 + f2 > c, then the radius of the circle is real
If g2 + f2 = 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 g2 + f2 < c, then the radius of the circle is imaginary. Such a circle have a real center and an imaginary radius
How to Find the Equation of a Circle
Let us solve some problems to find the equation of a circle.
Solved Examples
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 = r2, here (h, k) = (3, 5), r = 6 => (x-3)2 + (y-5)2 = 62
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 x2 + y2 = r2 Thus, for the given condition, the equation of a circle is given as x2 + y2 = 92
The equation of the given circle is x2 + y2 − 16x − 12y + 19 = 0. Find the center and radius of the circle.
Solution:
The given equation is in the form of x2 + y2 + 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 = √g2 + f2 – 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 x2 + y2 − 4x − 6y − 4 = 0 in standard form
Solution:
The given equation of the circle, x2 + y2 − 4x − 6y − 4 = 0 Rearranging the above equation we get, (x2 − 4x) + (y2 − 6y) − 4 = 0 => (x2 − 4x) + (y2 − 6y) = 4 Completing the square for x => (x2 – 4x + (-2)2) + (y2 − 6y) = 4 + (-2)2 Similarly, completing the square for y => (x2 – 4x + (-2)2) + (y2 − 6y + (-3)2) = 4 + (-2)2 + (-3)2 => (x2 – 4x + 4) + (y2 − 6y + 9) => (x – 2) 2 + (y – 3) 2 This is the standard form of the equation of the given circle.
