The Key Principle: Radius $\perp$ Tangent#
A tangent is a line that touches a circle at exactly one point. The single most important fact — from Euclidean Geometry — is:
The radius drawn to the point of tangency is perpendicular to the tangent line.
This means: if you know the gradient of the radius, the tangent gradient is the negative reciprocal.
$$m_{\text{tan}} = -\frac{1}{m_{\text{rad}}}$$This one fact is the engine behind every tangent question in Analytical Geometry.
1. The 4-Step Method#
To find the equation of a tangent to a circle at a given point:
| Step | Action | Tool |
|---|---|---|
| 1 | Find the centre of the circle | Read from standard form, or complete the square |
| 2 | Calculate $m_{\text{rad}}$ | Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ using centre and contact point |
| 3 | Find $m_{\text{tan}}$ | $m_{\text{tan}} = -\frac{1}{m_{\text{rad}}}$ (perpendicular gradients) |
| 4 | Write the tangent equation | Point-gradient form: $y - y_1 = m(x - x_1)$ using $m_{\text{tan}}$ and the contact point |
Worked Example 1 — Basic Tangent#
Find the equation of the tangent to $(x - 1)^2 + (y + 2)^2 = 25$ at the point $(4; 2)$.
Step 1 — Centre: $(1; -2)$
Step 2 — Gradient of radius:
$$m_{\text{rad}} = \frac{2 - (-2)}{4 - 1} = \frac{4}{3}$$Step 3 — Gradient of tangent:
$$m_{\text{tan}} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$$Step 4 — Equation:
$$y - 2 = -\frac{3}{4}(x - 4)$$$$y = -\frac{3}{4}x + 3 + 2$$$$\boxed{y = -\frac{3}{4}x + 5}$$Check: Verify $(4; 2)$ is on the circle: $(4-1)^2 + (2+2)^2 = 9 + 16 = 25\;\checkmark$
Worked Example 2 — Tangent at the Origin#
Find the equation of the tangent to $x^2 + y^2 = 20$ at $(2; -4)$.
Step 1 — Centre: $(0; 0)$
Step 2:
$$m_{\text{rad}} = \frac{-4 - 0}{2 - 0} = -2$$Step 3:
$$m_{\text{tan}} = -\frac{1}{-2} = \frac{1}{2}$$Step 4:
$$y - (-4) = \frac{1}{2}(x - 2)$$$$y + 4 = \frac{1}{2}x - 1$$$$\boxed{y = \frac{1}{2}x - 5}$$Check: Point on circle: $4 + 16 = 20\;\checkmark$
Worked Example 3 — Completing the Square First#
Find the equation of the tangent to $x^2 + y^2 - 6x + 4y - 12 = 0$ at $(7; 1)$.
Step 1 — Complete the square to find the centre:
$$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$$$$(x - 3)^2 + (y + 2)^2 = 25$$Centre: $(3; -2)$, radius $= 5$
Step 2:
$$m_{\text{rad}} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}$$Step 3:
$$m_{\text{tan}} = -\frac{4}{3}$$Step 4:
$$y - 1 = -\frac{4}{3}(x - 7)$$$$y = -\frac{4}{3}x + \frac{28}{3} + 1$$$$\boxed{y = -\frac{4}{3}x + \frac{31}{3}}$$2. Special Case: Horizontal and Vertical Tangents#
When the Radius is Vertical ($m_{\text{rad}}$ is undefined)#
If the contact point is directly above or below the centre (same $x$-coordinate), the radius is vertical and the tangent is horizontal: $y = k$ where $k$ is the $y$-coordinate of the contact point.
When the Radius is Horizontal ($m_{\text{rad}} = 0$)#
If the contact point is directly left or right of the centre (same $y$-coordinate), the radius is horizontal and the tangent is vertical: $x = k$ where $k$ is the $x$-coordinate of the contact point.
Worked Example 4 — Horizontal Tangent#
Find the tangent to $(x - 2)^2 + (y - 3)^2 = 9$ at $(2; 6)$.
Centre: $(2; 3)$. The contact point has the same $x$-coordinate as the centre.
The radius is vertical → the tangent is horizontal:
$$\boxed{y = 6}$$3. Tangent from an External Point#
A harder exam scenario: you’re given an external point and asked to find the tangent(s) to the circle through that point.
The Strategy#
An external point has two tangent lines to the circle. Use the condition that $\Delta = 0$ (the line touches the circle at exactly one point) or the perpendicular distance from centre to line equals the radius.
Worked Example 5 — Tangent from External Point#
Find the equations of the tangents to $x^2 + y^2 = 5$ from the point $(3; -1)$.
Let the tangent have the form $y - (-1) = m(x - 3)$, i.e., $y = mx - 3m - 1$.
Substitute into the circle equation:
$$x^2 + (mx - 3m - 1)^2 = 5$$Expand and simplify:
$$x^2 + m^2x^2 - 2mx(3m + 1) + (3m + 1)^2 = 5$$$$(1 + m^2)x^2 - 2m(3m + 1)x + (3m + 1)^2 - 5 = 0$$For a tangent, $\Delta = 0$:
$$[2m(3m+1)]^2 - 4(1+m^2)[(3m+1)^2 - 5] = 0$$$$4m^2(3m+1)^2 - 4(1+m^2)(9m^2 + 6m + 1 - 5) = 0$$$$m^2(3m+1)^2 - (1+m^2)(9m^2 + 6m - 4) = 0$$Expanding both sides:
$$9m^4 + 6m^3 + m^2 - (9m^2 + 6m - 4 + 9m^4 + 6m^3 - 4m^2) = 0$$$$9m^4 + 6m^3 + m^2 - 9m^4 - 6m^3 + 4m^2 - 9m^2 - 6m + 4 = 0$$$$-4m^2 - 6m + 4 = 0$$$$2m^2 + 3m - 2 = 0$$$$(2m - 1)(m + 2) = 0$$$$m = \frac{1}{2} \quad \text{or} \quad m = -2$$Tangent 1: $y = \frac{1}{2}x - \frac{3}{2} - 1 = \frac{1}{2}x - \frac{5}{2}$
Tangent 2: $y = -2x + 6 - 1 = -2x + 5$
$$\boxed{y = \frac{1}{2}x - \frac{5}{2} \quad \text{and} \quad y = -2x + 5}$$4. Length of a Tangent from an External Point#
If $P(x_1; y_1)$ is outside a circle with centre $C(a; b)$ and radius $r$, the length of the tangent from $P$ to the circle is:
$$\ell = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}$$Why? By Pythagoras in the right triangle formed by the centre, the contact point, and the external point:
$$PC^2 = \ell^2 + r^2 \quad \Rightarrow \quad \ell^2 = PC^2 - r^2$$Worked Example 6 — Length of Tangent#
$$PC^2 = (5 - 1)^2 + (7 - 3)^2 = 16 + 16 = 32$$$$\ell = \sqrt{32 - 9} = \sqrt{23}$$Find the length of the tangent from $P(5; 7)$ to the circle $(x - 1)^2 + (y - 3)^2 = 9$.
5. Verifying a Tangent Using the Discriminant#
An alternative approach: substitute the line into the circle equation. If $\Delta = 0$, the line is a tangent.
Worked Example 7 — Verifying a Tangent#
Show that $y = 2x + 5$ is a tangent to $x^2 + y^2 = 5$.
Substitute:
$$x^2 + (2x + 5)^2 = 5$$$$x^2 + 4x^2 + 20x + 25 = 5$$$$5x^2 + 20x + 20 = 0$$$$x^2 + 4x + 4 = 0$$$$\Delta = 16 - 16 = 0\;\checkmark$$Since $\Delta = 0$, the line touches the circle at exactly one point — it is a tangent.
Point of contact: $x^2 + 4x + 4 = (x + 2)^2 = 0$, so $x = -2$, $y = 2(-2) + 5 = 1$.
Contact point: $(-2; 1)$.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Wrong gradient formula order | $m = \frac{y_2 - y_1}{x_2 - x_1}$, not the other way | Always: $\frac{\Delta y}{\Delta x}$ |
| Forgetting the negative reciprocal | $m_{\text{tan}} = -\frac{1}{m_{\text{rad}}}$, not $\frac{1}{m_{\text{rad}}}$ | Perpendicular means negative and reciprocal |
| Using centre coordinates in the line equation | The tangent passes through the contact point, not the centre | Use the contact point $(x_1; y_1)$ in $y - y_1 = m(x - x_1)$ |
| Not verifying the point is on the circle | If the point isn’t on the circle, the “tangent” is meaningless | Always check by substituting the point into the circle equation |
| Undefined gradient panic | If $m_{\text{rad}}$ is undefined (vertical), the tangent is simply $y = k$ | Handle horizontal/vertical tangents as special cases |
💡 Pro Tips for Exams#
1. Always Verify First#
Before starting the 4-step method, check that the given point is actually on the circle by substituting into the equation. If it’s not on the circle, it might be an external-point tangent problem instead.
2. The Discriminant Shortcut#
If the question says “show that the line is a tangent,” substitute the line into the circle equation and show $\Delta = 0$. This is often faster than the perpendicular-gradient method and is exactly what markers expect.
3. Two Methods, Same Answer#
You can always cross-check your tangent equation by:
- Finding it via the gradient method (Steps 1–4)
- Substituting it into the circle equation and confirming $\Delta = 0$
If both agree, you’re correct.