Equation of a Circle — Notes
1. Standard form
The circle with centre \((a,b)\) and radius \(r\) has equation
\(
(x-a)^2 + (y-b)^2 = r^2.
\)
Centre = \((a,b)\). Radius = \(r\).
Example: \((x-2)^2 + (y+3)^2 = 16\) → centre \((2,-3)\), radius \(4\).
2. General form
A circle can be written as
\(
x^2 + y^2 + 2gx + 2fy + c = 0.
\)
Then the centre is \((-g,-f)\) and the radius is \(\sqrt{g^2+f^2-c}\).
Converting to standard form (complete the square)
Group \(x\)-terms and \(y\)-terms, complete squares and move the constant:
Given: \( x^2 + y^2 - 4x + 6y - 12 = 0 \)
Group: \( (x^2 - 4x) + (y^2 + 6y) = 12 \)
Complete: \( (x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9 \)
Result: \( (x-2)^2 + (y+3)^2 = 25 \) → centre \((2,-3)\), radius \(5\).
3. Circle with given centre and radius
Directly substitute into standard form: centre \((a,b)\), radius \(r\):
\((x-a)^2 + (y-b)^2 = r^2\).
4. Circle with diameter endpoints
If \(A(x_1,y_1)\) and \(B(x_2,y_2)\) are ends of a diameter, one convenient form is
\(
(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0,
\)
or equivalently the circle with centre at the midpoint of \(AB\) and radius half the distance \(AB\).
5. Finding centre and radius — summary steps
- Write the equation as a polynomial: \(x^2 + y^2 + 2gx + 2fy + c = 0\) (if necessary divide through by a constant).
- Complete the square for \(x\) and \(y\): add and subtract the required constants.
- Rearrange into \((x-a)^2 + (y-b)^2 = r^2\).
- Read off the centre \((a,b)\) and radius \(r\).
6. Special cases & cautions
- If \(g^2 + f^2 - c < 0\) then there is no real circle (radius would be imaginary).
- If radius \(= 0\) the circle reduces to a single point (the centre).
- If the equation has coefficients multiplied (e.g. \(2x^2+2y^2+\dots=0\)), divide by that factor first.
7. Quick reference table
| Form | Equation | Centre | Radius |
|---|
| Standard |
\((x-a)^2+(y-b)^2=r^2\) |
\((a,b)\) |
\(r\) |
| General |
\(x^2+y^2+2gx+2fy+c=0\) |
\((-g,-f)\) |
\(\sqrt{g^2+f^2-c}\) |
| Diameter AB |
\((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\) |
midpoint of \(A,B\) |
half distance \(AB\) |
8. Examples (worked)
Example 1: \(x^2+y^2-6x+8y+10=0\)
\(
(x^2-6x)+(y^2+8y)=-10
\)
\(
(x-3)^2+(y+4)^2 = 15
\)
Centre \((3,-4)\), radius \(\sqrt{15}\).
Example 2: \(x^2+y^2-x-10y+5=0\)
\(
(x^2-x)+(y^2-10y)=-5
\)
\(
(x-\tfrac{1}{2})^2+(y-5)^2 = \tfrac{81}{4}
\)
Centre \((\tfrac{1}{2},5)\), radius \(\tfrac{9}{2}\).
Example 3 (scale present): \(2x^2+2y^2-8x+2y+2=0\)
Divide by 2:
\(
x^2+y^2-4x+y+1=0
\)
Complete squares:
\(
(x-2)^2+(y+\tfrac{1}{2})^2=\tfrac{13}{4}
\)
Centre \((2,-\tfrac{1}{2})\), radius \(\tfrac{\sqrt{13}}{2}\).