What’s New in Grade 11 Analytical Geometry?#
Grade 10 gave you three tools: distance, midpoint, and gradient. Grade 11 builds on all three and adds powerful new ideas:
- Angle of inclination — the link between gradient and trigonometry
- Angle between two lines — using inclination to find the acute angle where lines meet
- The equation of a circle — a new curve to work with algebraically
- Tangent lines to circles — combining the perpendicular gradient rule with circle geometry
1. Angle of Inclination#
The angle of inclination ($\theta$) is the angle a line makes with the positive x-axis, measured anti-clockwise.
$$\boxed{m = \tan\theta}$$This means: gradient IS the tangent of the inclination angle.
| Gradient | Inclination | Direction |
|---|---|---|
| $m > 0$ | $0° < \theta < 90°$ | Line slopes upward to the right |
| $m < 0$ | $90° < \theta < 180°$ | Line slopes downward to the right |
| $m = 0$ | $\theta = 0°$ | Horizontal line |
| $m$ undefined | $\theta = 90°$ | Vertical line |
Calculating $\theta$#
Step 1: Find the gradient $m$.
Step 2: Calculate $\theta = \tan^{-1}(m)$.
Step 3: If $m < 0$, your calculator gives a negative angle. Add $180°$ to get the actual inclination:
$$\theta = \tan^{-1}(m) + 180° \quad \text{(when } m < 0\text{)}$$Why add 180° for negative gradients?#
The inclination is always between $0°$ and $180°$. When $m < 0$, the line slopes downward, so $\theta$ is obtuse (between $90°$ and $180°$). Your calculator’s $\tan^{-1}$ function returns a value between $-90°$ and $90°$, so for negative gradients it gives a negative angle. Adding $180°$ shifts it into the correct range.
Worked Example 1#
Find the angle of inclination of the line through $A(1;\; 3)$ and $B(4;\; -3)$.
$m = \frac{-3 - 3}{4 - 1} = \frac{-6}{3} = -2$
Since $m < 0$: $\theta = \tan^{-1}(-2) + 180° = -63.43° + 180° = 116.57°$
Worked Example 2#
Find the angle of inclination of the line $y = 3x - 5$.
$m = 3 > 0$, so $\theta = \tan^{-1}(3) = 71.57°$
2. Angle Between Two Lines#
The acute angle between two lines can be found using their inclination angles:
$$\theta = |\theta_1 - \theta_2|$$If this gives an obtuse angle (> 90°), subtract from 180° to get the acute angle.
The Formula Method#
You can also use the formula directly:
$$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$This gives the tangent of the acute angle between the lines.
⚠️ This formula fails when $m_1 m_2 = -1$ (perpendicular lines). In that case, the angle is exactly $90°$.
Worked Example 3#
Find the acute angle between the lines with gradients $m_1 = 3$ and $m_2 = -1$.
Method 1 — Using inclination angles:
$\theta_1 = \tan^{-1}(3) = 71.57°$
$\theta_2 = \tan^{-1}(-1) + 180° = -45° + 180° = 135°$
$\theta = |135° - 71.57°| = 63.43°$
Since $63.43° < 90°$, this IS the acute angle. ✓
Method 2 — Using the formula:
$\tan\theta = \left|\frac{3 - (-1)}{1 + (3)(-1)}\right| = \left|\frac{4}{-2}\right| = 2$
$\theta = \tan^{-1}(2) = 63.43°$ ✓
Worked Example 4#
Find the angle between $y = 2x + 1$ and $y = -\frac{1}{3}x + 4$.
$m_1 = 2$, $m_2 = -\frac{1}{3}$
$\tan\theta = \left|\frac{2 - (-\frac{1}{3})}{1 + (2)(-\frac{1}{3})}\right| = \left|\frac{\frac{7}{3}}{\frac{1}{3}}\right| = 7$
$\theta = \tan^{-1}(7) = 81.87°$
3. The Equation of a Circle#
Standard Form#
$$\boxed{(x - a)^2 + (y - b)^2 = r^2}$$| Feature | How to read it |
|---|---|
| Centre | $(a;\; b)$ — the values being subtracted from $x$ and $y$ |
| Radius | $r = \sqrt{r^2}$ |
Special case: Centre at the origin → $x^2 + y^2 = r^2$
Writing the Equation#
Worked Example 5#
Write the equation of a circle with centre $(3;\; -2)$ and radius 5.
$(x - 3)^2 + (y - (-2))^2 = 5^2$
$$\boxed{(x - 3)^2 + (y + 2)^2 = 25}$$Worked Example 6#
Write the equation of a circle with centre $(-1;\; 4)$ and passing through $(2;\; 0)$.
First find $r$: $r^2 = (2 - (-1))^2 + (0 - 4)^2 = 9 + 16 = 25$
$$(x + 1)^2 + (y - 4)^2 = 25$$Reading the Equation#
Worked Example 7#
Find the centre and radius of $(x + 1)^2 + (y - 4)^2 = 16$.
$(x - (-1))^2 + (y - 4)^2 = 16$
Centre: $(-1;\; 4)$, Radius: $\sqrt{16} = 4$
⚠️ The sign trap: $(x + 1)$ means the centre’s x-coordinate is $-1$ (the opposite sign). $(y - 4)$ means the centre’s y-coordinate is $+4$.
4. Completing the Square — Converting to Standard Form#
When a circle equation is given in expanded form ($x^2 + y^2 + Dx + Ey + F = 0$), you must complete the square to find the centre and radius.
The Method#
- Group $x$-terms and $y$-terms: $(x^2 + Dx) + (y^2 + Ey) = -F$
- Complete the square for each group: add $\left(\frac{D}{2}\right)^2$ and $\left(\frac{E}{2}\right)^2$ to both sides
- Write in standard form and read off centre and radius
Worked Example 8#
Find the centre and radius: $x^2 + y^2 - 6x + 4y - 3 = 0$
Step 1 — Group:
$(x^2 - 6x) + (y^2 + 4y) = 3$
Step 2 — Complete the square:
Half of $-6$ is $-3$, and $(-3)^2 = 9$. Half of $4$ is $2$, and $2^2 = 4$.
$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4$
Step 3 — Standard form:
$(x - 3)^2 + (y + 2)^2 = 16$
Centre: $(3;\; -2)$, Radius: $4$
Worked Example 9#
Find the centre and radius: $x^2 + y^2 + 8x - 2y + 8 = 0$
$(x^2 + 8x) + (y^2 - 2y) = -8$
$(x^2 + 8x + 16) + (y^2 - 2y + 1) = -8 + 16 + 1$
$(x + 4)^2 + (y - 1)^2 = 9$
Centre: $(-4;\; 1)$, Radius: $3$
Worked Example 10#
Determine whether $x^2 + y^2 + 6x - 4y + 15 = 0$ is a valid circle.
$(x^2 + 6x + 9) + (y^2 - 4y + 4) = -15 + 9 + 4$
$(x + 3)^2 + (y - 2)^2 = -2$
$r^2 = -2 < 0$ → This is NOT a valid circle (radius squared cannot be negative).
5. Point Position — Inside, On, or Outside?#
Substitute the point into the left side of $(x - a)^2 + (y - b)^2$:
| Result | Position |
|---|---|
| $< r^2$ | Inside the circle |
| $= r^2$ | On the circle |
| $> r^2$ | Outside the circle |
Worked Example 11#
Is the point $(1;\; 2)$ inside, on, or outside $(x - 3)^2 + (y + 1)^2 = 25$?
$(1 - 3)^2 + (2 + 1)^2 = 4 + 9 = 13$
$13 < 25$ → the point is inside the circle.
Worked Example 12#
Determine the position of $(5;\; 3)$ relative to $x^2 + y^2 = 20$.
$5^2 + 3^2 = 25 + 9 = 34$
$34 > 20$ → the point is outside the circle.
6. Tangent to a Circle#
A tangent touches the circle at exactly one point and is perpendicular to the radius at that point.
The 3-Step Method#
Step 1: Find the gradient of the radius (from centre to tangent point): $m_r$
Step 2: Tangent gradient = negative reciprocal: $m_t = -\frac{1}{m_r}$
Step 3: Use point-gradient form with the tangent point: $y - y_1 = m_t(x - x_1)$
Worked Example 13#
Find the equation of the tangent to $x^2 + y^2 = 25$ at $(3;\; 4)$.
Step 1: Centre = $(0;\; 0)$. $m_r = \frac{4 - 0}{3 - 0} = \frac{4}{3}$
Step 2: $m_t = -\frac{3}{4}$
Step 3: $y - 4 = -\frac{3}{4}(x - 3)$
$y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4}$
Verify: $m_r \times m_t = \frac{4}{3} \times -\frac{3}{4} = -1$ ✓ (perpendicular)
Worked Example 14 — Shifted Centre#
Find the tangent to $(x - 2)^2 + (y + 1)^2 = 20$ at $(6;\; 1)$.
Step 1 — Verify point is on circle: $(6-2)^2 + (1+1)^2 = 16 + 4 = 20$ ✓
Step 2: Centre: $(2;\; -1)$. $m_r = \frac{1 - (-1)}{6 - 2} = \frac{2}{4} = \frac{1}{2}$
Step 3: $m_t = -2$
Step 4: $y - 1 = -2(x - 6)$
$$y = -2x + 13$$Worked Example 15 — Horizontal and Vertical Tangents#
Find the tangent to $x^2 + y^2 = 25$ at $(5;\; 0)$.
Centre = $(0;\; 0)$. $m_r = \frac{0}{5} = 0$ (horizontal radius).
A horizontal radius means the tangent is vertical: $x = 5$.
Find the tangent at $(0;\; 5)$.
$m_r = \frac{5}{0}$ = undefined (vertical radius).
A vertical radius means the tangent is horizontal: $y = 5$.
7. Finding the Equation of a Circle Given Conditions#
Given: Centre and a Tangent Line#
Worked Example 16#
A circle has centre $(3;\; 1)$ and is tangent to the x-axis. Find its equation.
The distance from the centre to the x-axis = the radius = $|1| = 1$
$$(x - 3)^2 + (y - 1)^2 = 1$$Given: Endpoints of a Diameter#
Worked Example 17#
A and B are endpoints of a diameter. $A(-2;\; 3)$ and $B(4;\; -1)$. Find the equation.
Centre = midpoint of AB: $\left(\frac{-2+4}{2};\; \frac{3+(-1)}{2}\right) = (1;\; 1)$
Radius = half the diameter = $\frac{1}{2}\sqrt{(4-(-2))^2 + (-1-3)^2} = \frac{1}{2}\sqrt{36+16} = \frac{1}{2}\sqrt{52} = \sqrt{13}$
$$(x - 1)^2 + (y - 1)^2 = 13$$🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Forgetting $+180°$ for negative gradients | $\tan^{-1}(m)$ gives a negative angle when $m < 0$; inclination must be $0° \leq \theta < 180°$ | Always add $180°$ when $m < 0$ |
| Signs in the circle equation | $(x + 1)^2$ means centre x-coordinate is $-1$, not $+1$ | The sign you see is the opposite of the centre coordinate |
| Completing the square — forgetting both sides | Adding 9 and 4 to the left but not the right changes the equation | Whatever you add to the left, add to the right too |
| Tangent gradient = reciprocal (not negative reciprocal) | The tangent is perpendicular to the radius, so $m_t = -\frac{1}{m_r}$ | Always negate AND reciprocal |
| Not verifying the point is on the circle | If the point isn’t on the circle, you can’t find a tangent there | Substitute before calculating — the result must equal $r^2$ |
| Using the formula for angle between lines when $m_1 m_2 = -1$ | The denominator becomes zero → undefined | If $m_1 m_2 = -1$, the lines are perpendicular, so $\theta = 90°$ |
| Saying $r^2 < 0$ is a circle | A circle requires $r^2 > 0$ | If completing the square gives $r^2 \leq 0$, state that it’s not a valid circle |
| Forgetting special tangent cases | When the radius is horizontal or vertical, the tangent is vertical or horizontal | If $m_r = 0$, tangent is $x = x_1$; if $m_r$ is undefined, tangent is $y = y_1$ |
💡 Pro Tips for Exams#
1. The Perpendicular Check#
After finding a tangent, verify: $m_r \times m_t = -1$. This 5-second check catches most gradient errors.
2. Completing the Square — The Quick Method#
For $x^2 + y^2 + Dx + Ey + F = 0$:
- Centre: $\left(-\frac{D}{2};\; -\frac{E}{2}\right)$
- $r^2 = \left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F$
This skips the completing-the-square steps entirely.
3. Angle of Inclination Decision Tree#
- Find $m$
- If $m \geq 0$: $\theta = \tan^{-1}(m)$ → done
- If $m < 0$: $\theta = \tan^{-1}(m) + 180°$ → done
4. Circle + Tangent Questions — The Exam Pattern#
Most exam questions follow this sequence:
- Complete the square to find the centre and radius
- Show a point is on the circle (substitute and verify = $r^2$)
- Find the tangent equation at that point
- Find the angle of inclination of the tangent
Practice this 4-step chain until it’s automatic.
🔗 Related Grade 11 topics:
- Trig Ratios & Reduction — $\tan\theta$ gives the angle of inclination; reduction handles obtuse angles
- Circle Geometry — the geometric approach to circles complements the algebraic approach here
- Quadratic Equations — completing the square converts expanded form to standard form
📌 Grade 10 foundation: Core Formulas — distance, midpoint, gradient
📌 Grade 12 extension: Circle Equation and Tangents — external tangents, length of tangent