Equation of a Tangent to a Circle Revision Notes for Grade 12 NSC Matric Mathematics

Equation of a Tangent to a Circle

What is a tangent to a circle?

A tangent is a straight line that touches the circumference of a circle at exactly one point. This single point where the tangent meets the circle is called the point of tangency.

The diagram above shows a circle with centre $C(a; b)$ and radius $r$ . The straight line $AB$ is tangent to the circle at point $D(x; y)$ .

Key properties of tangents

The perpendicular property

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Critical Concept: Perpendicular Relationship

The most important property of tangents is that the radius drawn to the point of tangency is always perpendicular to the tangent line.

This means:

$CD \perp AB$ (the radius $CD$ is perpendicular to tangent $AB$ )
The angle between the radius and tangent is always $90°$
$\angle CDA = \angle CDB = 90°$

The gradient relationship

When two lines are perpendicular, the product of their gradients equals $-1$ .

For a radius and its tangent: $m_{\text{radius}} \times m_{\text{tangent}} = -1$

This relationship is essential for finding tangent equations.

Method for finding the equation of a tangent

Follow these five steps to find the equation of a tangent to a circle:

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Step-by-Step Method for Finding Tangent Equations

Step 1: Write the circle equation in standard form

Convert the circle equation to the form $(x - a)^2 + (y - b)^2 = r^2$

Use completing the square if necessary.

Step 2: Identify the centre coordinates

From the standard form, the centre is at $(a; b)$ .

Step 3: Find the gradient of the radius

Use the gradient formula: $m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1}$

Calculate from the centre to the point of tangency.

Step 4: Calculate the gradient of the tangent

Since the radius and tangent are perpendicular: $m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}$

Step 5: Write the tangent equation

Use the point-slope form: $y - y_1 = m(x - x_1)$

Substitute the tangent gradient and the coordinates of the point of tangency.

Worked example 1: Tangent at a given point

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Worked Example: Finding Tangent at Given Point

Question: Find the equation of the tangent to the circle $x^2 + y^2 - 2y + 6x - 7 = 0$ at point $F(-2; 5)$ .

Solution:

Step 1: Complete the square to get standard form

$x^2 + y^2 - 2y + 6x - 7 = 0$
$x^2 + 6x + y^2 - 2y = 7$
$(x^2 + 6x + 9) + (y^2 - 2y + 1) = 7 + 9 + 1$
$(x + 3)^2 + (y - 1)^2 = 17$

Step 2: The centre is $C(-3; 1)$ and radius is $\sqrt{17}$

Step 3: Find gradient of radius $CF$

$m_{CF} = \frac{5 - 1}{-2 - (-3)} = \frac{4}{1} = 4$

Step 4: Find gradient of tangent

Since $CF \perp$ tangent: $m_{\text{tangent}} = -\frac{1}{4}$

Step 5: Write the tangent equation

$y - 5 = -\frac{1}{4}(x - (-2))$
$y - 5 = -\frac{1}{4}(x + 2)$
$y = -\frac{1}{4}x + \frac{9}{2}$

Answer: The equation of the tangent is $y = -\frac{1}{4}x + \frac{9}{2}$

Worked example 2: Finding intersection points first

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Worked Example: Tangents at Intersection Points

Question: The line $y = x + 4$ cuts the circle $x^2 + y^2 = 26$ at points $P$ and $Q$ . Find the equations of the tangents at $P$ and $Q$ .

Solution:

Step 1: Find intersection points

Substitute $y = x + 4$ into $x^2 + y^2 = 26$ :

$x^2 + (x + 4)^2 = 26$
$x^2 + x^2 + 8x + 16 = 26$
$2x^2 + 8x - 10 = 0$
$x^2 + 4x - 5 = 0$
$(x - 1)(x + 5) = 0$

So $x = 1$ or $x = -5$

When $x = 1$ : $y = 1 + 4 = 5$ , giving $P(1; 5)$

When $x = -5$ : $y = -5 + 4 = -1$ , giving $Q(-5; -1)$

Step 2: Find tangent at $P(1; 5)$

Centre is $O(0; 0)$ , so gradient of $OP = \frac{5 - 0}{1 - 0} = 5$

Gradient of tangent $= -\frac{1}{5}$

Using $y - y_1 = m(x - x_1)$ : $y - 5 = -\frac{1}{5}(x - 1)$

Simplifying: $y = -\frac{1}{5}x + \frac{26}{5}$

Step 3: Find tangent at $Q(-5; -1)$

Gradient of $OQ = \frac{-1 - 0}{-5 - 0} = \frac{1}{5}$

Gradient of tangent $= -5$

Using $y - y_1 = m(x - x_1)$ : $y - (-1) = -5(x - (-5))$

Simplifying: $y = -5x - 26$

Worked example 3: Parallel tangents

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Worked Example: Finding Parallel Tangents

Question: Find the equations of tangents to the circle $x^2 + (y - 1)^2 = 80$ that are parallel to the line $y = \frac{1}{2}x + 1$ .

Solution:

Since the tangents are parallel to $y = \frac{1}{2}x + 1$ , they have gradient $m = \frac{1}{2}$ .

The centre is at $(0; 1)$ and radius is $\sqrt{80}$ .

For tangents with gradient $\frac{1}{2}$ , the radius must have gradient $-2$ (since $\frac{1}{2} \times (-2) = -1$ ).

Points where the radius has gradient $-2$ :

From centre $(0; 1)$ to point $(x; y)$ on the circle: $\frac{y - 1}{x - 0} = -2$

So $y - 1 = -2x$ , giving $y = -2x + 1$

Substitute into circle equation:

$x^2 + (-2x + 1 - 1)^2 = 80$
$x^2 + (-2x)^2 = 80$
$x^2 + 4x^2 = 80$
$5x^2 = 80$
$x^2 = 16$
$x = \pm 4$

When $x = 4$ : $y = -2(4) + 1 = -7$ , giving point $A(4; -7)$

When $x = -4$ : $y = -2(-4) + 1 = 9$ , giving point $B(-4; 9)$

The tangent equations are:

At $A$ : $y + 7 = \frac{1}{2}(x - 4) \rightarrow y = \frac{1}{2}x - 9$
At $B$ : $y - 9 = \frac{1}{2}(x + 4) \rightarrow y = \frac{1}{2}x + 11$

Common exam tips

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Essential Exam Success Tips

Always convert the circle equation to standard form $(x - a)^2 + (y - b)^2 = r^2$ first
Draw a sketch to visualise the problem - it helps avoid errors
Remember the perpendicular relationship - this is the key to all tangent problems
Check your gradient calculation - many errors occur in this step
Be careful with signs when finding the centre coordinates from the equation

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Key Points to Remember:

A tangent touches a circle at exactly one point
The radius to the point of tangency is always perpendicular to the tangent
For perpendicular lines: $m_1 \times m_2 = -1$
Convert circle equations to standard form: $(x - a)^2 + (y - b)^2 = r^2$
Use point-slope form $y - y_1 = m(x - x_1)$ for the final tangent equation

Equation of a Tangent to a Circle (Grade 12 NSC Matric Mathematics): Revision Notes