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URL: https://nakafa.com/en/subjects/mathematics/circle/circle-and-tangent-line
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/circle/circle-and-tangent-line/en.mdx
Learn tangent lines to circles, including equations, properties, external points, and common tangents.
---
## Definition of Circle Tangent Line
A circle tangent line is a line that intersects the circle at exactly one point. The intersection point between the tangent line and the circle is called the point of tangency.
Component: LineEquation
Props:
- title: Circle Tangent Line
- description: A tangent line intersects the circle at exactly one point.
- 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: [
{ x: 0, y: 0, z: 0 },
{ x: 3, y: 0, z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "O", at: 0, offset: [-0.5, -0.5, 0] },
{ text: "P", at: 1, offset: [0.5, 0, 0] },
],
},
{
points: [
{ x: 3, y: -3, z: 0 },
{ x: 3, y: 3, z: 0 },
],
color: getColor("CYAN"),
showPoints: false,
labels: [{ text: "Tangent line", at: 1, offset: [1.5, 0, 0] }],
},
]
- cameraPosition: [0, 0, 10]
- showZAxis: false
**Important property:** A tangent line is always perpendicular to the radius of the circle at the point of tangency.
## Equation of Circle Tangent Line
### Tangent Line Through a Point on the Circle
If point $$(x_1, y_1)$$ lies on the circle $$x^2 + y^2 = r^2$$, then the equation of the tangent line at that point is:
Visible text: If point lies on the circle , then the equation of the tangent line at that point is:
```math
x_1 \cdot x + y_1 \cdot y = r^2
```
For a circle with center $$(a, b)$$:
Visible text: For a circle with center :
```math
(x_1 - a)(x - a) + (y_1 - b)(y - b) = r^2
```
### Tangent Line with a Given Gradient
The equation of the tangent line to circle $$x^2 + y^2 = r^2$$ with gradient $$m$$ is:
Visible text: The equation of the tangent line to circle with gradient is:
```math
y = mx \pm r\sqrt{1 + m^2}
```
For a circle with center $$(a, b)$$:
Visible text: For a circle with center :
```math
y - b = m(x - a) \pm r\sqrt{1 + m^2}
```
## Tangent Lines from an External Point
From a point outside the circle, two tangent lines can be drawn to the circle.
Component: LineEquation
Props:
- title: Two Tangent Lines from External Point
- description: From point $$P$$ outside the circle, two tangent lines can
be drawn.
Visible text: From point outside the circle, two tangent lines can
be drawn.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 2.5 * Math.cos(angle),
y: 2.5 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [{ x: 0, y: 0, z: 0 }],
color: getColor("ORANGE"),
showPoints: true,
labels: [{ text: "O", at: 0, offset: [-0.5, -0.5, 0] }],
},
{
points: [{ x: 5, y: 0, z: 0 }],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "P", at: 0, offset: [0.5, 0, 0] }],
},
{
points: (() => {
const P = { x: 5, y: 0 };
const O = { x: 0, y: 0 };
const r = 2.5;
const d = Math.sqrt((P.x - O.x) ** 2 + (P.y - O.y) ** 2);
// Angle from center to external point
const theta = Math.atan2(P.y - O.y, P.x - O.x);
// Angle of tangent line from center
const alpha = Math.asin(r / d);
// First tangent point
const T1x = O.x + r * Math.cos(theta + Math.PI / 2 - alpha);
const T1y = O.y + r * Math.sin(theta + Math.PI / 2 - alpha);
return [
{ x: P.x, y: P.y, z: 0 },
{ x: T1x, y: T1y, z: 0 },
];
})(),
color: getColor("TEAL"),
showPoints: true,
labels: [{ text: "T₁", at: 1, offset: [0.3, 0.5, 0] }],
},
{
points: (() => {
const P = { x: 5, y: 0 };
const O = { x: 0, y: 0 };
const r = 2.5;
const d = M ... [truncated; 1729 chars]
- cameraPosition: [0, 0, 10]
- showZAxis: false
### Length of Tangent Line
If $$P(x_1, y_1)$$ is a point outside the circle with center $$O(a, b)$$ and radius $$r$$, then the length of the tangent line from P to the circle is:
Visible text: If is a point outside the circle with center and radius , then the length of the tangent line from P to the circle is:
```math
PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}
```
## Common Tangent Lines of Two Circles
### External Common Tangent Lines
External common tangent lines are lines that touch both circles and do not intersect the line connecting the two circle centers.
Component: LineEquation
Props:
- title: External Common Tangent Lines
- description: Lines that touch both circles from the same side.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: -3 + 1.5 * Math.cos(angle),
y: 1.5 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 3 + 2 * Math.cos(angle),
y: 2 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: -3, y: 0, z: 0 },
{ x: 3, y: 0, z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "O₁", at: 0, offset: [-0.5, -0.5, 0] },
{ text: "O₂", at: 1, offset: [0.5, -0.5, 0] },
],
},
{
points: (() => {
const O1 = { x: -3, y: 0 };
const O2 = { x: 3, y: 0 };
const r1 = 1.5;
const r2 = 2;
const d = Math.sqrt((O2.x - O1.x) ** 2 + (O2.y - O1.y) ** 2);
// Angle between center line and common tangent
const alpha = Math.asin((r2 - r1) / d);
// Angle from O1 to O2
const theta = Math.atan2(O2.y - O1.y, O2.x - O1.x);
// Tangent points on first circle
const T1x = O1.x + r1 * Math.cos(theta + Math.PI/2 - alpha);
const T1y = O1.y + r1 * Math.sin(theta + Math.PI/2 - alpha);
// Tangent points on second circle
const T2x = O2.x + r2 * Math.cos(thet ... [truncated; 2700 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false
Length of external common tangent line:
```math
l = \sqrt{d^2 - (r_1 - r_2)^2}
```
where $$d$$ is the distance between the two circle centers.
Visible text: where is the distance between the two circle centers.
### Internal Common Tangent Lines
Internal common tangent lines are lines that touch both circles and intersect the line connecting the two circle centers.
Component: LineEquation
Props:
- title: Internal Common Tangent Lines
- description: Lines that touch both circles from opposite sides.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: -3 + 1.5 * Math.cos(angle),
y: 1.5 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 3 + 1.5 * Math.cos(angle),
y: 1.5 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: -3, y: 0, z: 0 },
{ x: 3, y: 0, z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "O₁", at: 0, offset: [-0.5, -0.5, 0] },
{ text: "O₂", at: 1, offset: [0.5, -0.5, 0] },
],
},
{
points: (() => {
const O1 = { x: -3, y: 0 };
const O2 = { x: 3, y: 0 };
const r1 = 1.5;
const r2 = 1.5;
const d = Math.sqrt((O2.x - O1.x) ** 2 + (O2.y - O1.y) ** 2);
// For internal common tangent, we need to find the angle
const alpha = Math.asin((r1 + r2) / d);
// Angle from O1 to O2
const theta = Math.atan2(O2.y - O1.y, O2.x - O1.x);
// For internal tangent, tangent points are on opposite sides
// First circle: top side
const T1x = O1.x + r1 * Math.cos(theta + Math.PI/2 - alpha);
const T1y = O1.y + r1 * Math.sin(theta + Math.PI/2 - alpha);
// ... [truncated; 2937 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false
Length of internal common tangent line:
```math
l = \sqrt{d^2 - (r_1 + r_2)^2}
```
## Determining Tangent Line Equations
### Determining Tangent Line Through a Point on the Circle
Find the equation of the tangent line to circle $$x^2 + y^2 = 25$$ at point $$(3, 4)$$.
Visible text: Find the equation of the tangent line to circle at point .
**Solution:**
Since point $$(3, 4)$$ lies on the circle (can be verified: $$3^2 + 4^2 = 9 + 16 = 25$$), the equation of the tangent line is:
Visible text: Since point lies on the circle (can be verified: ), the equation of the tangent line is:
Component: MathContainer
Children:
```math
x_1 \cdot x + y_1 \cdot y = r^2
```
```math
3x + 4y = 25
```
### Determining Tangent Line with a Given Gradient
Find the equation of the tangent line to circle $$x^2 + y^2 = 16$$ that is parallel to line $$3x - 4y + 5 = 0$$.
Visible text: Find the equation of the tangent line to circle that is parallel to line .
**Solution:**
The gradient of line $$3x - 4y + 5 = 0$$ is $$m = \frac{3}{4}$$.
Visible text: The gradient of line is .
Equation of tangent line with gradient $$m = \frac{3}{4}$$:
Visible text: Equation of tangent line with gradient :
Component: MathContainer
Children:
```math
y = mx \pm r\sqrt{1 + m^2}
```
```math
y = \frac{3}{4}x \pm 4\sqrt{1 + \left(\frac{3}{4}\right)^2}
```
```math
y = \frac{3}{4}x \pm 4\sqrt{1 + \frac{9}{16}}
```
```math
y = \frac{3}{4}x \pm 4\sqrt{\frac{25}{16}}
```
```math
y = \frac{3}{4}x \pm 4 \cdot \frac{5}{4}
```
```math
y = \frac{3}{4}x \pm 5
```
Therefore, the equations of the tangent lines are:
- $$y = \frac{3}{4}x + 5$$ or $$3x - 4y + 20 = 0$$
- $$y = \frac{3}{4}x - 5$$ or $$3x - 4y - 20 = 0$$
Visible text: - or
- or
### Calculating the Length of Tangent Line from External Point
Find the length of the tangent line from point $$P(7, 1)$$ to circle $$x^2 + y^2 = 25$$.
Visible text: Find the length of the tangent line from point to circle .
**Solution:**
Circle center $$O(0, 0)$$ and radius $$r = 5$$.
Visible text: Circle center and radius .
Component: MathContainer
Children:
```math
PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}
```
```math
PT = \sqrt{(7 - 0)^2 + (1 - 0)^2 - 25}
```
```math
PT = \sqrt{49 + 1 - 25}
```
```math
PT = \sqrt{25}
```
```math
PT = 5
```
## Practice Problems
1. Find the equation of the tangent line to circle $$x^2 + y^2 = 36$$ at point $$(-3, 3\sqrt{3})$$!
2. Find the equation of the tangent line to circle $$(x - 2)^2 + (y + 3)^2 = 25$$ that is perpendicular to line $$2x + y - 5 = 0$$!
3. From point $$A(10, 0)$$ tangent lines are drawn to circle $$x^2 + y^2 = 36$$. Find:
- Length of tangent line
- Coordinates of tangent points
4. Two circles are centered at $$O_1(-4, 0)$$ with radius $$2$$ and $$O_2(4, 0)$$ with radius $$3$$. Find the length of the external common tangent line!
5. Find the equation of the tangent line to circle $$x^2 + y^2 - 6x + 4y - 12 = 0$$ that passes through point $$(8, -2)$$!
Visible text: 1. Find the equation of the tangent line to circle at point !
2. Find the equation of the tangent line to circle that is perpendicular to line !
3. From point tangent lines are drawn to circle . Find:
- Length of tangent line
- Coordinates of tangent points
4. Two circles are centered at with radius and with radius . Find the length of the external common tangent line!
5. Find the equation of the tangent line to circle that passes through point !
### Answer Key
1. **Tangent line equation at a point on the circle**
Verify point on circle:
```math
(-3)^2 + (3\sqrt{3})^2 = 9 + 27 = 36 \space \checkmark
```
Tangent line equation:
```math
x_1 \cdot x + y_1 \cdot y = r^2
```
```math
-3x + 3\sqrt{3}y = 36
```
```math
-x + \sqrt{3}y = 12
```
```math
x - \sqrt{3}y + 12 = 0
```
2. **Tangent line perpendicular to a given line**
Gradient of line $$2x + y - 5 = 0$$ is $$m_1 = -2$$.
Since perpendicular, then $$m_2 = \frac{1}{2}$$.
Tangent line equation:
```math
y - b = m(x - a) \pm r\sqrt{1 + m^2}
```
```math
y + 3 = \frac{1}{2}(x - 2) \pm 5\sqrt{1 + \frac{1}{4}}
```
```math
y + 3 = \frac{1}{2}(x - 2) \pm 5 \cdot \frac{\sqrt{5}}{2}
```
```math
y + 3 = \frac{1}{2}x - 1 \pm \frac{5\sqrt{5}}{2}
```
Therefore: $$x - 2y - 4 \pm 5\sqrt{5} = 0$$
3. **Tangent line from external point**
- Length of tangent line:
```math
AT = \sqrt{10^2 + 0^2 - 36} = \sqrt{100 - 36} = \sqrt{64} = 8
```
- Coordinates of tangent points can be found using the tangent line equation from external point.
4. **External common tangent line**
```math
d = \sqrt{(4-(-4))^2 + (0-0)^2} = 8
```
```math
l = \sqrt{d^2 - (r_2 - r_1)^2} = \sqrt{64 - 1} = \sqrt{63} = 3\sqrt{7}
```
5. **Tangent line through external point**
Circle: $$(x - 3)^2 + (y + 2)^2 = 25$$
Center $$(3, -2)$$, radius $$r = 5$$
Verify point $$(8, -2)$$ outside circle:
```math
(8-3)^2 + (-2+2)^2 = 25 + 0 = 25
```
Point is exactly on the circle! Therefore the tangent line equation is:
```math
(8-3)(x-3) + (-2+2)(y+2) = 25
```
```math
5(x-3) + 0 = 25
```
```math
x - 3 = 5
```
```math
x = 8
```
Visible text: 1. **Tangent line equation at a point on the circle**
Verify point on circle:
Tangent line equation:
2. **Tangent line perpendicular to a given line**
Gradient of line is .
Since perpendicular, then .
Tangent line equation:
Therefore:
3. **Tangent line from external point**
- Length of tangent line:
- Coordinates of tangent points can be found using the tangent line equation from external point.
4. **External common tangent line**
5. **Tangent line through external point**
Circle:
Center , radius
Verify point outside circle:
Point is exactly on the circle! Therefore the tangent line equation is: