Circle equation formula refers to the equation of a circle which represents the centre-radius form of the circle. To recall, a circle is referred to a round shape boundary where all the points on the boundary are equidistant from the centre. An equation is generally required to represent the circle. There are basically two forms of representation:
- Standard Form
- General Form
Standard Form of Circle Equation
| (x – a)2 + (y – b)2 = r2 |
Where,
- (a, b) are the coordinates of centre
- r is the radius
General form of Circle Equation
| x2 + y2 + Ax + By + C = 0 |
Solved Examples
Question 1: If the centre point and radius of a circle is given as (4, 5) and 7 respectively. Represent this as a circle equation.
Solution:
Given parameters are:
Center (a, b) = (4, 5);
Radius r = 7
The standard form of circle equation is,
(x − a)2 + (y − b)2 = r2
So, (x − 4)2 + (y − 5)2 = 72
So, (x − 4)2 + (y − 5)2 = 49
Question 2: Find the centre and radius of the circle whose equation is given by x2 + y2 – 10x + 14y + 38 = 0.
Solution:
Given circle equation is:
x2 + y2 – 10x + 14y + 38 = 0
x2 -2(5)x + y2 + 2(7)y + 38 = 0
This can also be written as:
x2 -2(5)x + (5)2 + y2 + 2(7)y + (7)2 + 38 – 25 – 49 = 0
(x – 5)2 + (y + 7)2 – 36 = 0
(x – 5)2 + (y + 7)2 = 36
(x – 5)2 + (y + 7)2 = 62
Comparing this with the standard form (x – a)2 + (y – b)2 = r2
a = 5, b = -7 and r = 6
Therefore, centre = (5, -7) and radius = 6 units.
Question 3: Write the general form of the circle equation with centre (2, 3) and radius 1 unit.
Solution:
Given,
Centre = (a, b) = (2, 3)
Radius = r = 1
Standard form of circle equation is (x – a)2 + (y – b)2 = r2
Substituting the values of centre and radius,
(x – 2)2 + (y – 3)2 = 12
x2 – 4x + 4 + y2 – 6y + 9 = 1
x2 + y2 – 4x – 6y + 10 – 1 = 0
x2 + y2 – 4x – 6y + 9 = 0
This of the form x2 + y2 + Ax + By + C = 0 where A = -4, B = -6, C = 9
Hence, the general form of the circle equation is x2 + y2 – 4x – 6y + 9 = 0.
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