Equation of a Circle

We will learn how to find the equation of a circle whose centre and radius are given.

Case I: If the centre and radius of a circle be given, we can determine its equation:

To find the equation of the circle whose centre is at the origin O and radius r units:

Let M (x, y) be any point on the circumference of the required circle. 

Therefore, the locus of the moving point M = OM = radius of the circle = r 

OM\(^{2}\) = r\(^{2}\)

x\(^{2}\) + y\(^{2}\) = r\(^{2}\), which is the required equation of the circle.

 

Case II: To find the equation of the circle whose centre is at C (h, k) and radius r units:

Let M (x, y) be any point on the circumference of the requited circle. Therefore, the locus of the moving point M = CM = radius of the circle = r

CM\(^{2}\) = r\(^{2}\)

(x - h)\(^{2}\) + (y - k)\(^{2}\) = r\(^{2}\), which is the required equation of the circle.

 

Note: 

(i) The above equation is known as the central from of the equation of a circle.

(ii) Referred to O as pole and OX as initial line of polar co-ordinate system, if the polar co-ordinates of M be (r, θ) then we shall have,

r = OM = radius of the circle = a and ∠MOX = θ.

Then, from the above figure we get,

x = ON = a cos θ and y = MN = a sin θ

Here, x = a cos θ and y = a sin θ represent the parametric equations of the circle x\(^{2}\) + y\(^{2}\) = r\(^{2}\).


Solved examples to find the equation of a circle:

1. Find the equation of a circle whose centre is (4, 7) and radius 5.

Solution:

The equation of the required circle is

(x - 4)\(^{2}\) + (y - 7)\(^{2}\) = 5\(^{2}\)

x\(^{2}\) - 16x + 16 + y\(^{2}\) - 14y + 49 = 25

x\(^{2}\) + y\(^{2}\) - 16x - 14y + 40 = 0


2. Find the equation of a circle whose radius is 13 and the centre is at the origin.

Solution:

The equation of the required circle is

x\(^{2}\) + y\(^{2}\) = 13\(^{2}\)

x\(^{2}\) + y\(^{2}\) = 169

 The Circle





11 and 12 Grade Math 

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