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URL: https://nakafa.com/en/subjects/mathematics/analytic-geometry/equation-of-circle
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Learn circle equations from (x-h)² + (y-k)² = r² to general form. Learn finding center, radius, and converting between forms with practice problems.
---
## Understanding Circle Equations
Circle equation is a mathematical formula that describes all points forming a circle on a coordinate plane. Imagine you have a compass and want to draw a circle on coordinate paper. Well, the circle equation tells us which coordinates the compass pencil tip will pass through.
Why is this useful? Because by knowing the circle equation, we can immediately tell where the **center of the circle** is and what its **radius** is without having to draw the circle first.
## Circle Centered at Origin
Let's start with the easiest case first: a circle whose center is at the origin $$O(0,0)$$.
Visible text: Let's start with the easiest case first: a circle whose center is at the origin .
If we have a circle with center at $$O(0,0)$$ and radius $$r$$, then every point $$(x,y)$$ on that circle has the same distance from the center, which is $$r$$.
Visible text: If we have a circle with center at and radius , then every point on that circle has the same distance from the center, which is .
Using the distance formula, we get:
```math
\sqrt{x^2 + y^2} = r
```
If we square both sides, we get the circle equation with center at origin:
```math
x^2 + y^2 = r^2
```
Let's visualize this first.
Component: LineEquation
Props:
- title: Circle with Center at Origin
- description: Circle with center $$O(0,0)$$ and radius $$3$$.
Visible text: Circle with center and radius .
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 3 * Math.cos(angle),
y: 3 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: Array.from({ length: 2 }, (_, i) => {
const center = { x: 0, y: 0 };
const radius = 3;
const angle = 0;
return {
x: center.x + i * radius * Math.cos(angle),
y: center.y + i * radius * Math.sin(angle),
z: 0,
};
}),
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "O(0, 0)", at: 0, offset: [-0.5, -0.5, 0] },
{ text: "r = 3", at: 1, offset: [0, -0.5, 0] },
],
},
{
points: Array.from({ length: 2 }, (_, i) => {
const xMin = -4;
const xMax = 4;
const y = 0;
return {
x: xMin + i * (xMax - xMin),
y: y,
z: 0,
};
}),
color: getColor("AMBER"),
showPoints: false,
smooth: false,
},
{
points: Array.from({ length: 2 }, (_, i) => {
const yMin = -4;
const yMax = 4;
const x = 0;
return {
x: x,
y: yMin + i * (yMax - yMin),
z: 0,
};
}),
color: getColor("AMBER"),
showPoints: false,
smooth: false,
},
]
- cameraPosition: [0, 0, 10]
- showZAxis: false
For the circle above, the equation is $$x^2 + y^2 = 9$$ because the radius is $$3$$, so $$r^2 = 3^2 = 9$$.
Visible text: For the circle above, the equation is because the radius is , so .
## Circle with Arbitrary Center
Now what if the center isn't at the origin? Say the circle center is at point $$P(a,b)$$ with radius $$r$$.
Visible text: Now what if the center isn't at the origin? Say the circle center is at point with radius .
Every point $$(x,y)$$ on this circle must have the same distance $$r$$ from center $$P(a,b)$$. Using the distance formula:
Visible text: Every point on this circle must have the same distance from center . Using the distance formula:
```math
\sqrt{(x-a)^2 + (y-b)^2} = r
```
After squaring, we get the **general circle equation**:
```math
(x-a)^2 + (y-b)^2 = r^2
```
If we visualize this, it will look like this:
Component: LineEquation
Props:
- title: Circle with Arbitrary Center
- description: Circle with center $$P(2,-1)$$ and radius $$2$$.
Visible text: Circle with center and radius .
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 2 + 2 * Math.cos(angle),
y: -1 + 2 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: Array.from({ length: 2 }, (_, i) => {
const center = { x: 2, y: -1 };
const radius = 2;
const angle = 0;
return {
x: center.x + i * radius * Math.cos(angle),
y: center.y + i * radius * Math.sin(angle),
z: 0,
};
}),
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "P(2, -1)", at: 0, offset: [-0.5, 0.5, 0] },
{ text: "r = 2", at: 1, offset: [0, -0.5, 0] },
],
},
{
points: Array.from({ length: 2 }, (_, i) => {
const xMin = -1;
const xMax = 5;
const y = 0;
return {
x: xMin + i * (xMax - xMin),
y: y,
z: 0,
};
}),
color: getColor("AMBER"),
showPoints: false,
smooth: false,
},
{
points: Array.from({ length: 2 }, (_, i) => {
const yMin = -4;
const yMax = 2;
const x = 0;
return {
x: x,
y: yMin + i * (yMax - yMin),
z: 0,
};
}),
color: getColor("AMBER"),
showPoints: false,
smooth: false,
},
]
- cameraPosition: [0, 0, 10]
- showZAxis: false
The equation of the circle above is $$(x-2)^2 + (y+1)^2 = 4$$ because the center is $$P(2,-1)$$ and the radius is $$2$$, so $$r^2 = 2^2 = 4$$.
Visible text: The equation of the circle above is because the center is and the radius is , so .
## General Form of Circle Equation
Sometimes we find circle equations that have already been expanded into general form. For example, from the equation $$(x-1)^2 + (y-2)^2 = 9$$, if we expand it:
Visible text: Sometimes we find circle equations that have already been expanded into general form. For example, from the equation , if we expand it:
Component: MathContainer
Children:
```math
(x-1)^2 + (y-2)^2 = 9
```
```math
x^2 - 2x + 1 + y^2 - 4y + 4 = 9
```
```math
x^2 + y^2 - 2x - 4y + 5 = 9
```
```math
x^2 + y^2 - 2x - 4y - 4 = 0
```
This last form is called the **general form of circle equation**:
```math
x^2 + y^2 + Dx + Ey + F = 0
```
If we have an equation in general form, we can convert it back to standard form using **completing the square** technique.
> Not all equations of the form $$x^2 + y^2 + Dx + Ey + F = 0$$ are circle equations. The condition is $$D^2 + E^2 - 4F > 0$$. If this value is zero, then it's just a single point, and if negative, then there's no curve at all.
Visible text: > Not all equations of the form are circle equations. The condition is . If this value is zero, then it's just a single point, and if negative, then there's no curve at all.
## Determining Center and Radius
From the standard form $$(x-a)^2 + (y-b)^2 = r^2$$, we can directly know:
Visible text: From the standard form , we can directly know:
The center is at $$(a,b)$$ and the radius is $$r$$ (obtained from $$r = \sqrt{r^2}$$).
Visible text: The center is at and the radius is (obtained from ).
While from the general form $$x^2 + y^2 + Dx + Ey + F = 0$$, we can determine:
Visible text: While from the general form , we can determine:
Component: MathContainer
Children:
```math
\text{Center: } \left(-\frac{D}{2}, -\frac{E}{2}\right)
```
```math
\text{Radius: } r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}
```
## Practice Problems
1. Determine the equation of a circle centered at $$(3,-2)$$ with radius $$5$$.
2. Given a circle with equation $$x^2 + y^2 - 6x + 4y - 12 = 0$$. Determine the center and radius of the circle.
3. A circle passes through point $$(6,-8)$$ and is centered at $$O(0,0)$$. Determine the equation of the circle.
4. Determine the equation of a circle that has a diameter with endpoints at $$A(1,3)$$ and $$B(7,-1)$$.
Visible text: 1. Determine the equation of a circle centered at with radius .
2. Given a circle with equation . Determine the center and radius of the circle.
3. A circle passes through point and is centered at . Determine the equation of the circle.
4. Determine the equation of a circle that has a diameter with endpoints at and .
### Answer Key
1. **Solution**:
Given center $$(a,b) = (3,-2)$$ and radius $$r = 5$$.
Using the circle equation formula:
```math
(x-a)^2 + (y-b)^2 = r^2
```
```math
(x-3)^2 + (y-(-2))^2 = 5^2
```
```math
(x-3)^2 + (y+2)^2 = 25
```
So the circle equation is $$(x-3)^2 + (y+2)^2 = 25$$.
2. **Solution**:
From the equation $$x^2 + y^2 - 6x + 4y - 12 = 0$$, we identify the coefficients:
$$D = -6$$, $$E = 4$$, $$F = -12$$
```math
\text{Center: } \left(-\frac{D}{2}, -\frac{E}{2}\right) = \left(-\frac{(-6)}{2}, -\frac{4}{2}\right) = (3, -2)
```
```math
r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F} = \frac{1}{2}\sqrt{(-6)^2 + 4^2 - 4(-12)}
```
```math
r = \frac{1}{2}\sqrt{36 + 16 + 48} = \frac{1}{2}\sqrt{100} = \frac{1}{2} \times 10 = 5
```
So the circle center is at $$(3,-2)$$ with radius $$5$$.
3. **Solution**:
Since the circle is centered at $$O(0,0)$$ and passes through point $$(6,-8)$$, the radius is the distance from the center to that point.
```math
r = \sqrt{(6-0)^2 + (-8-0)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10
```
The circle equation: $$x^2 + y^2 = r^2 = 10^2 = 100$$
4. **Solution**:
The circle center is the midpoint of diameter $$AB$$:
```math
\text{Center} = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{1+7}{2}, \frac{3+(-1)}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1)
```
The radius is half the diameter length:
```math
\text{Length } AB = \sqrt{(x_B-x_A)^2 + (y_B-y_A)^2} = \sqrt{(7-1)^2 + (-1-3)^2}
```
```math
= \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}
```
```math
r = \frac{\text{diameter}}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13}
```
The circle equation: $$(x-4)^2 + (y-1)^2 = (\sqrt{13})^2 = 13$$
Visible text: 1. **Solution**:
Given center and radius .
Using the circle equation formula:
So the circle equation is .
2. **Solution**:
From the equation , we identify the coefficients:
, ,
So the circle center is at with radius .
3. **Solution**:
Since the circle is centered at and passes through point , the radius is the distance from the center to that point.
The circle equation:
4. **Solution**:
The circle center is the midpoint of diameter :
The radius is half the diameter length:
The circle equation: