Equation of a Circle

Equation of a Circle

Equation of a circle is a second degree equation in x and y which satisfies all the points on the circumference of a circle in a Cartesian Plane. If r is the length of radius and (a, b) is the centre of a circle, then this circle has an equation (x – a)² + (y – b)² = r². 

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An equation of the form x² + y² + 2gx + 2fy + c = 0 represents a circle if g² + f² – c > 0, and is then an equation of the circle with centre at (–g, –f) and radius √(g² + f² – c).

Equation of a Circle With Centre at Origin

Let O(0, 0) be the centre of a circle. If P(x, y) be any point on the circumference of the circle, then OP = r.

By using distance formula,

                             OP² = (x – 0)² + (y – 0)²

                ∴          x² + y² = r²           … … … (i)

The relation (i) is satisfied by every point on the circle, hence it represents the equation of circle with centre at origin and radius r.

Example 1: Find the equation of circle with centre at origin and radius 5 units.

Solution:

                Here, radius (r) = 5 units

                The equation of a circle with centre at origin is

                              x² + y² = r²

                or,          x² + y² = 5²

                ∴            x² + y² = 25 is the required equation.

Equation of Circle With Centre at (h, k)

Let C(h, k) be the centre and r be the radius of a circle. If P(x, y) be any point on the circle, then CP = r.

By using distance formula,

                              CP² = (x – h)² + (y – k)²

                or,          (x – h)² + (y – k)² = r²      … … … (i)

This relation is true for every point on the circle. Hence, represents a circle with centre at (h, k) and radius r.

Equation (i) can be written as

                              x²–2hx+h²+y²–2ky+k²–r² = 0

                or,          x²+y²–2hx–2ky+h²+k²–r² = 0

                or,          x² + y² + 2gx + 2fy + c = 0               … … … (ii)

                where g = -h, f = -k and c = h² + k² – r².

The equation (ii) is said to be the General equation of circle. This equation of circle has the following characteristics.

a.       It is second degree in x and y.

b.      Coefficient of x² = coefficient of y²

c.       Coefficient of xy = 0.

Thus the general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 will represent a circle if a = b and h = 0.

Example 2: Find the equation of a circle whose centre is at point (-4, 3) and radius 4 units.

Solution:

                Here, centre of circle is (-4, 3)

                i.e. h = -4, k = 3 and radius (r) = 4.

                The equation of circle with centre:(h, k) and radius=r is

                              (x–h)²+(y – k)²= r²

                or,          (x + 4)² + (y – 3)² = 4²

                or,          x² + 8x +16 + y² – 6y + 9 = 16

∴             x² + y² + 8x – 6y + 9 = 0 is the required equation. 

              

Example 3: Find the centre and radius of the circle x² + y² – 6x + 4y – 23 = 0

Solution:

                The given equation is

                              x²+y²–6x+4y–23 = 0

                or,          x² – 6x + y² + 4y – 23 = 0

                or,          x² – 2.x.3 + 3² + y² + 2.y.2 + 2² = 23 + 3² + 2²

                or,          (x – 3)² + (y + 2)² = 23 + 9 + 4

                or,          (x – 3)² + (y + 2)² = 6²

Comparing it with (x – h)² + (y – k)² = r², we have h = 3, k = -2 and r = 6

Hence, centre of circle is (h, k) = (3, - 2) and radius r = 6.

Equation of Circle in a Diameter Form

Let O be the centre and A(x₁, y₁) and B(x₂, y₂) be the ends of diameter AB of the circle. If P(x, y) be any point on the circumference of the circle, then

Since, ∠APB is the angle in the semi-circle ∠APB = 90°.

∴             m₁ × m₂ = -1

or,          (y – y₁)(y – y₂) = - (x –x₁)(x – x₂)

or,          (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0

Hence, it is the equation of circle in diameter form.

Example 4: Find the equation of a circle whose ends of the diameter are at (-1, 3) and (5, 7).

Solution:

                Here, ends of the diameter of the circle are (-1, 3) and (5, 7).

                                i.e. x₁ = -1, y₁ = 3, x₂ = 5 and y₂ = 7.

                The equation of circle in diameter form is

                              (x–x₁)(x–x₂) + (y–y₁)(y–y₂) = 0

                or,          (x + 1)(x – 5) + (y – 3)(y – 7) = 0

     or,          x² – 5x + x – 5 + y² –  7y – 3y + 21 = 0

                or,          x² + y² – 4x – 10y + 16 = 0 is the required equation.

Example 5: Find the equation of a circle with centre (-3, 6) and touching the x-axis.

Solution:

                Here, centre of circle is (-3, 6). As the circle touches the x-axis, radius of the circle is 6.

                Then the equation of the circle is

                              (x–h)² + (y–k)² = r²

                or,          (x + 3)² + (y – 6)² = 6²

                or,          x² + 6x + 9 + y² – 12y + 36 = 36

                ∴           x² +y² +6x -12y + 9 = 0 is the required equation or circle. 

You can comment your questions or problems regarding the equation of circle here.