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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