A circle is defined as the locus of all the points in a plane, which are at a fixed distance from a fixed point. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. The radius of a circle is denoted by the letter ‘r’ or ‘R’. In this article, we shall study, the equation of a circle in the centre radius form.
Standard Equation of Circle:
![Centre Radius Form](https://thefactfactor.com/wp-content/uploads/2021/02/Centre-Radius-Form-01.png)
By distance formula
![Centre Radius Form](https://thefactfactor.com/wp-content/uploads/2021/02/Centre-Radius-Form-02.png)
Let P(x, y) be any point on a circle having centre at O(0, 0) and radius ‘a’.
Then, OP = radius of a circle = a
By distance formula
As this contains less number of constants, the equation is called the standard equation of a circle.
Centre Radius Form:
![Centre Radius Form](https://thefactfactor.com/wp-content/uploads/2021/02/Centre-Radius-Form-03.png)
By distance formula
![blank](https://thefactfactor.com/wp-content/plugins/wp-fastest-cache-premium/pro/images/blank.gif)
Let P(x, y) be any point on a circle having centre at C(h, k) and radius ‘r’.
Then, CP = radius of a circle = r
By distance formula
This form of the equation of a circle is called centre radius form.
Equation of Circle When Centre and Radius are Given
Algorithm:
- Write given centre ≡ (h, k) and given radius = r
- Use centre radius form (x – h)2 + (y – k)2 = r2
- Simplify the equation and write it in standard form ax2 + by2 + 2gx + 2fy + c = 0
Example – 01:
- Find the equation of circle having centre at (2, -3) and radius 5.
- Solution:
Given centre (2, -3) ≡ (h, k) and radius = r = 5
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 2)2 + (y + 3)2 = 52
∴ x2 – 4x + 4 + y2 + 6y + 9 = 25
∴ x2 + y2 – 4x + 6y + 13 – 25 = 0
∴ x2 + y2 – 4x + 6y – 12 = 0
Ans: The required equation of circle is x2 + y2 – 4x + 6y – 12 = 0
Example – 02:
- Find equation of circle having centre at origin and radius 4.
- Solution:
Given centre (0, 0) ≡ (h, k) and radius = r = 4
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 0)2 + (y + 0)2 = 42
∴ x2 + y2 = 16
Ans: The required equation of circle is x2 + y2 = 16
Example – 03:
- Find the equation of circle having centre at (3, -2) and radius 5.
- Solution:
Given centre (3, -2) ≡ (h, k) and radius = r = 5
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 3)2 + (y + 2)2 = 52
∴ x2 – 6x + 9 + y2 + 4y + 4 = 25
∴ x2 + y2 – 6x + 4y + 13 – 25 = 0
∴ x2 + y2 – 6x + 4y – 12 = 0
Ans: The required equation of circle is x2 + y2 – 6x + 4y – 12 = 0
Example – 04:
- Find the equation of circle having centre at (-3, -2) and radius 6.
- Solution:
Given centre (-3, -2) ≡ (h, k) and radius = r = 6
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x + 3)2 + (y + 2)2 = 62
∴ x2 + 6x + 9 + y2 + 4y + 4 = 36
∴ x2 + y2 + 6x + 4y + 13 – 36 = 0
∴ x2 + y2 + 6x + 4y – 23 = 0
Ans: The required equation of circle is x2 + y2 + 6x + 4y – 23 = 0
Example – 05:
- Find the equation of circle having centre at (a cosα, a sinα) and radius a.
- Solution:
Given centre (a cosα, a sinα) ≡ (h, k) and radius = r = 6
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
(x – a cosα)2 + (y – a sinα)2 = a2
∴ x2 – 2xa cosα + a2 cos2α + y2 – 2ya sinα + a2 sin2α = a2
∴ x2 + y2 – 2ax cosα – 2ay sinα + a2 cos2α + a2 sin2α = a2
∴ x2 + y2 – 2ax cosα – 2ay sinα + a2 (cos2α + sin2α) – a2 = 0
∴ x2 + y2 – 2ax cosα – 2ay sinα + a2 (1) – a2 = 0
∴ x2 + y2 – 2acosα.x – 2a sinα.y = 0
Ans: The required equation of circle is x2 + y2 – 2acosα.x – 2a sinα.y = 0
Example – 06:
- Find the equation of circle having centre at (a, b) and radius √a2+b2
- Solution:
Given centre (a, b) ≡ (h, k) and radius = r = √a2+b2
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – a)2 + (y – b)2 = (√a2+b2 )2
∴ x2 – 2ax + a2 + y2 – 2by + b2 = a2 + b2
∴ x2 + y2 – 2ax – 2by + a2 + b2 – a2 – b2 = 0
∴ x2 + y2 – 2ax – 2by = 0
Ans: The required equation of circle is x2 + y2 – 2ax – 2by = 0
Equation of Circle When Centre and Point on the Circle are Given:
Algorithm:
![blank](https://thefactfactor.com/wp-content/plugins/wp-fastest-cache-premium/pro/images/blank.gif)
- Write given centre C ≡ (h, k) and given point say P (x, y)
- Use distance formula to find CP represent it as radius = r
- Use centre radius form (x – h)2 + (y – k)2 = r2
- Simplify the equation and write it in standard form ax2 + by2 + 2gx + 2fy + c = 0
Example 07:
- Find the equation of circle having centre at (-3, 1) and passing through (5, 2)
- Solution:
Given centre C(-3, 1) ≡ (h, k) and point on circle P(5, 2)
By distance formula
CP2 = (5 + 3)2 + (2 – 1)2 = 64 + 1 = 65
∴ CP = √65
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x + 3)2 + (y – 1)2 = (√65)2
∴ x2 + 6x + 9 + y2 – 2y + 1 = 65
∴ x2 + y2 + 6x – 2y + 10 – 65 = 0
∴ x2 + y2 + 6x – 2y – 55 = 0
Ans: The required equation of circle is x2 + y2 + 6x – 2y – 55 = 0
Example 08:
- Find the equation of circle having centre at (2, -1) and passing through (3, 6)
- Solution:
Given centre C(2, -1) ≡ (h, k) and point on circle P(3, 6)
By distance formula
CP2 = (3 – 2)2 + (6 + 1)2 =1 + 49 = 50
∴ CP = √50
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 2)2 + (y + 1)2 = (√50)2
∴ x2 – 4x + 4 + y2 + 2y + 1 = 50
∴ x2 + y2 – 4x + 2y + 5 – 50 = 0
∴ x2 + y2 – 4x + 2y – 45 = 0
Ans: The required equation of circle is x2 + y2 – 4x + 2y – 45 = 0
Example 09:
- Find the equation of circle having centre at (2, -3) and passing through P(-3, 5)
- Solution:
Given centre C(2, – 3) ≡ (h, k) and point on circle (-3, 5)
By distance formula
CP2 = (-3 – 2)2 + (5 + 3)2 = 25 + 68 = 89
∴ CP = √89
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 2)2 + (y + 3)2 = (√89)2
∴ x2 – 4x + 4 + y2 + 6y + 9 = 89
∴ x2 + y2 – 4x + 6y + 13 – 89 = 0
∴ x2 + y2 – 4x + 6y – 76 = 0
Ans: The required equation of circle is x2 + y2 – 4x + 6y – 76 = 0
Example 10:
- Find equation of circle passing through (-3, 2) and having centre at the point of intersection of lines x – 2y = 4 and 2x + 5y + 1 = 0
- Solution:
Solving the equations x – 2y = 4 and 2x + 5y + 1 = 0 we get x = 2 and y = -1
Thus centre C(2, – 1) ≡ (h, k) and point on circle P(-3, 2)
By distance formula
CP2 = (-3 – 2)2 + (2 + 1)2 = 25 + 9 = 34
∴ CP = √34
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x – 2)2 + (y + 1)2 = (√34)2
∴ x2 – 4x + 4 + y2 + 2y + 1 = 34
∴ x2 + y2 – 4x + 2y + 5 – 34 = 0
∴ x2 + y2 – 4x + 2y – 29 = 0
Ans: The required equation of circle is x2 + y2 – 4x + 2y – 29 = 0
Example 11:
- Find equation of circle passing through intersection of x + 3y = 0 and 2x – 7y = 0 and having centre at the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0
- Solution:
Solving the equations x + 3y = 0 and 2x – 7y = 0 we get x = 0 and y = 0
Hence the circle passes through O(0, 0)
Solving the equations x + y + 1 = 0 and x – 2y + 4 = 0 we get x = -2 and y = 1
Hence the centre of circle is C(-2, 1)
Thus centre C(-2, 1) ≡ (h, k) and point on circle O(0, 0)
By distance formula
CO2 = (0 + 2)2 + (0 – 1)2 = 4 + 1 = 5
∴ CO = √5
By centre radius form, the equation of circle is given by
(x – h)2 + (y – k)2 = r2
∴ (x + 2)2 + (y – 1)2 = (√5)2
∴ x2 + 4x + 4 + y2 – 2y + 1 = 5
∴ x2 + y2 + 4x – 2y + 5 – 5 = 0
∴ x2 + y2 + 4x – 2y = 0
Ans: The required equation of circle is x2 + y2 + 4x – 2y = 0
Find equation of a circle having centre at (- 4, 0) and radius 5. Ans: x2 + y2 + 8x – 34 = 0
6. Find equation of a circle having centre at (7, 2) and radius 3. Ans: x2 + y2 – 14x – 4y + 44 = 0
7. Find equation of a circle having centre at (- 5, 1) and radius 2. Ans: x2 + y2 + 10x – 2y + 14 = 0
8. Find equation of a circle having centre at (1, -1) and passing through (3, 2) (M – 78)
Ans: x2 + y2 – 2x + 2y – 11 = 0
9. Find equation of a circle having centre at (2, -1) and passing through (5, 3) (O – 81)
Ans: x2 + y2 – 4x + 2y – 20 = 0.
10. Find equation of a circle having centre at (2, -3) and passing through (-1, 2) (M – 98)
Ans: x2 + y2 – 4x + 6y – 21 = 0
11. Find equation of a circle having centre at (1, -2) and passing through intersection of lines 3x + y = 14 and 2x + 5y = 18. Ans: x2 + y2 – 2x + 4y – 20 = 0
12. Find equation of a circle having centre at (- 4, 3) and Y – axis is tangent to it. (O – 78)
Ans: x2 + y2 + 8x – 6y + 9 = 0
13. Find equation of a circle having centre at (2, – 6) and touching X – axis.
Ans: x2 + y2 – 4x+ 12y + 4 = 0
14. Find equation of a circle having centre at (2, 3) and touching X – axis.
Ans: x2 + y2 – 4x – 6y + 4 = 0
15. Find equation of a circle having centre at (7, – 2) and touching X – axis.
Ans: x2 + y2 – 14x + 4y + 49 = 0
16. Find equation of a circle which touches X – axis at (-1, 0) and have radius 5.
Ans:
One reply on “Equation of Circle (Centre Radius Form)”
very good examples and clear explanations