Equation of a Tangent Line to a Circle

Understanding Tangent Line Equations

Have you ever noticed how a bicycle wheel moves on the road? When the wheel rotates, there's one point at the edge of the wheel that always touches the asphalt perfectly. Well, imagine if we could draw a straight line from that point of contact - that's what we call a tangent line!

This lesson focuses on finding the explicit mathematical equation of a tangent line. The equation tells us exactly which straight line touches the circle at one point.

The ability to determine tangent line equations is very useful in engineering, physics, and even art. Imagine an architect designing a bridge arch, or an engineer calculating projectile trajectories.

Substitution for Specific Points

When we already know which point on the circle will be touched by the tangent line, our work becomes more focused. Let's look at a systematic approach for this case.

Tangent Line at a Specific Point

Constructing the tangent line equation when the point of tangency is known.

For circle $x^2 + y^2 = r^2$ with point of tangency $M(p, q)$ , we use the direct substitution formula:

px + qy = r^2

Why does this formula work? Because the tangent line is always perpendicular to the radius at the point of tangency. If the radius has direction vector $(p, q)$ , then the tangent line has the same normal vector, which is $(p, q)$ .

For general circle $(x-h)^2 + (y-k)^2 = r^2$ with point of tangency $M(p, q)$ , the formula becomes:

(p-h)(x-h) + (q-k)(y-k) = r^2

The advantage of this method is speed and accuracy. We don't need to calculate gradients separately or perform complex algebraic manipulations.

Distance Approach for Given Gradient

Now we move to a more challenging case: finding the tangent line equation when only its gradient is known. Here we use the principle that the distance from the circle's center to the tangent line must equal the radius.

Tangent Line with Specific Slope

Using distance approach to find equation with given gradient.

Suppose we have circle $(x-h)^2 + (y-k)^2 = r^2$ and want to find a tangent line with gradient $m$ . The tangent line has form $y = mx + c$ .

The key is using the point-to-line distance formula. The distance from center $(h,k)$ to line $mx - y + c = 0$ is:

d = \frac{|mh - k + c|}{\sqrt{m^2 + 1}}

Since the line is tangent to the circle, this distance must exactly equal the radius:

\frac{|mh - k + c|}{\sqrt{m^2 + 1}} = r

Thus we obtain two constant values:

c = k - mh \pm r\sqrt{m^2 + 1}

Therefore, the equations of both tangent lines are:

y = mx + k - mh \pm r\sqrt{m^2 + 1}

Quadratic Method for External Points

The most challenging case is when we want to draw tangent lines from a point outside the circle. From one external point, we can draw exactly two tangent lines with different gradients.

Tangent Line from External Point

Using quadratic method in gradient for points outside the circle.

To solve this problem, we use the quadratic equation in gradient approach. Let the external point be $P(x_0, y_0)$ and the circle have equation $x^2 + y^2 = r^2$ .

The tangent line from point $P$ has form $y - y_0 = m(x - x_0)$ , or $y = mx - mx_0 + y_0$ .

Substitute into the circle equation and use the condition that discriminant equals zero:

x^2 + (mx - mx_0 + y_0)^2 = r^2

Discriminant equals zero gives a quadratic equation in $m$ :

(x_0^2 - r^2)m^2 - 2x_0y_0m + (y_0^2 - r^2) = 0

The two roots of this equation are the gradients of both tangent lines.

Complete Applied Example

Let's solve one concrete case to clarify understanding. Determine the tangent line equation for circle $x^2 + y^2 = 25$ passing through point $A(7, 1)$ .

Step $1$ : Verify that point $A$ is outside the circle.

7^2 + 1^2 = 49 + 1 = 50 > 25 \quad \checkmark

Step $2$ : Form quadratic equation in gradient with $x_0 = 7$ , $y_0 = 1$ , $r^2 = 25$ :

(49 - 25)m^2 - 2(7)(1)m + (1 - 25) = 0

Step $3$ : Solve using quadratic formula:

m = \frac{7 \pm \sqrt{49 + 576}}{24} = \frac{7 \pm \sqrt{625}}{24} = \frac{7 \pm 25}{24}

So $m_1 = \frac{32}{24} = \frac{4}{3}$ and $m_2 = \frac{-18}{24} = -\frac{3}{4}$ .

Step $4$ : Construct final equations:

y - 1 = \frac{4}{3}(x - 7) \Rightarrow 4x - 3y = 25

Practice

Determine the tangent line equation for circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ .
Find the tangent line equation for circle $x^2 + y^2 = 9$ with gradient $-\frac{1}{2}$ .
Determine the tangent line equation for circle $(x-3)^2 + (y-4)^2 = 25$ passing through point $(11, 10)$ .
A circle has equation $x^2 + y^2 - 6x - 8y = 0$ . Determine the tangent line equation parallel to line $3x + 4y = 7$ .

Answer Key

Solution:

For circle $x^2 + y^2 = 16$ at point $(2\sqrt{2}, 2\sqrt{2})$ , use the formula:

$x \cdot x_1 + y \cdot y_1 = r^2$

Step $1$ : Substitute tangent point coordinates
$x \cdot 2\sqrt{2} + y \cdot 2\sqrt{2} = 16$
Step $2$ : Simplify by dividing both sides by $2\sqrt{2}$

$x + y = \frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$

So the tangent line equation is $x + y = 4\sqrt{2}$ .
Solution:

For gradient $m = -\frac{1}{2}$ and $r = 3$ , use the formula:

$y = mx \pm r\sqrt{1 + m^2}$

Step $1$ : Calculate value of $\sqrt{1 + m^2}$

$\sqrt{1 + m^2} = \sqrt{1 + \left(-\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$

Step $2$ : Substitute into formula

$y = -\frac{1}{2}x \pm 3 \cdot \frac{\sqrt{5}}{2} = -\frac{1}{2}x \pm \frac{3\sqrt{5}}{2}$

So the two tangent line equations are: $y = -\frac{1}{2}x + \frac{3\sqrt{5}}{2}$ and $y = -\frac{1}{2}x - \frac{3\sqrt{5}}{2}$ .
Solution:

Circle $(x-3)^2 + (y-4)^2 = 25$ has center $(3,4)$ and radius $r = 5$ .

Step $1$ : Check position of point $(11,10)$

$(11-3)^2 + (10-4)^2 = 8^2 + 6^2 = 64 + 36 = 100 > 25$

Point is outside the circle.

Step $2$ : Use quadratic equation in gradient. With coordinate translation to circle center, point $(11,10)$ becomes $(8,6)$ relative to center.

Step $3$ : Quadratic equation in gradient for circle $x^2 + y^2 = 25$ from point $(8,6)$ :
$(8^2 - 25)m^2 - 2(8)(6)m + (6^2 - 25) = 0$
Step $4$ : Using quadratic formula:
$m = \frac{96 \pm \sqrt{96^2 - 4(39)(11)}}{2(39)} = \frac{96 \pm \sqrt{9216 - 1716}}{78}$
Step $5$ : Simplify gradients and determine tangent line equation through point $(11,10)$ :
$m_1 = \frac{96 + 50\sqrt{3}}{78} \text{ and} m_2 = \frac{96 - 50\sqrt{3}}{78}$
Both tangent line equations can be written in form $y - 10 = m(x - 11)$ .
Solution:

Step $1$ : Convert circle equation to standard form by completing the square
$x^2 + y^2 - 6x - 8y = 0$
Center $(3,4)$ , radius $r = 5$ .

Step $2$ : Line $3x + 4y = 7$ can be written as $y = -\frac{3}{4}x + \frac{7}{4}$

Parallel line gradient: $m = -\frac{3}{4}$

Step $3$ : For circle with center $(3,4)$ , parallel tangent line equation:
$y - 4 = -\frac{3}{4}(x - 3) \pm 5\sqrt{1 + \frac{9}{16}}$
Step $4$ : Simplify both equations
$y = -\frac{3}{4}x + \frac{50}{4} = -\frac{3}{4}x + \frac{25}{2}$
So the two tangent line equations are: $y = -\frac{3}{4}x + \frac{25}{2}$ and $y = -\frac{3}{4}x$ .