# Circle Equations, Lines, Tangents and Two Circles

> Additional Mathematics · CIE IGCSE 0606
> Source: https://www.owlsprep.com/study/cie-0606-u8-circle-equations-lines-tangents-and/

This guide covers all circle coordinate geometry content for CIE IGCSE Additional Maths 0606, including equation forms, line-circle intersections, tangents, and pairs of circles, aligned to the 2025-2027 syllabus.

**Prerequisites:** Quadratic equations and discriminant; Straight line coordinate geometry

## Learning objectives

- Derive and convert between the two standard forms of a circle's equation
- Classify line-circle intersections (chord/tangent/no meet) using the discriminant
- Find the equation of a tangent to a circle using the radius-tangent perpendicularity rule
- Classify the positional relationship between two circles and calculate their common chord equation

## Standard Forms of a Circle's Equation

**Centre-Radius Form** — For a circle with centre $(a,b)$ and radius $r$, the equation is derived from the distance formula between any point $(x,y)$ on the circle and its centre.

*Example:* A circle with centre $(2,-3)$ and radius 4 has equation $(x-2)^2 + (y+3)^2 = 16$

**Expanded General Form** — The expanded form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, with centre $(-g,-f)$ and radius $r=\sqrt{g^2+f^2-c}$ (the expression under the square root must be positive for a real circle).

*Example:* The circle $x^2+y^2-4x+6y-3=0$ has $g=-2, f=3, c=-3$, so centre $(2,-3)$ and radius $\sqrt{4+9+3}=4$

**Worked example:** Convert the expanded form equation $x^2 + y^2 + 2x - 4y - 4 = 0$ to centre-radius form, and state the centre and radius.

1. 1. Group x terms and y terms separately, move the constant to the right-hand side:

   $$(x^2 + 2x) + (y^2 - 4y) = 4$$
2. 2. Complete the square for x terms: $x^2+2x = (x+1)^2 -1$, and for y terms: $y^2-4y = (y-2)^2 -4$
3. 3. Substitute back and simplify:

   $$(x+1)^2 -1 + (y-2)^2 -4 = 4 \implies (x+1)^2 + (y-2)^2 = 9$$
4. 4. Identify centre $(-1, 2)$ and radius 3.

> **Exam tip:** Always check that the radius you calculate from the general form is positive: if $g^2 + f^2 - c$ is negative, the equation does not represent a real circle.

## Intersection of a Line and a Circle

To find the intersection of a line and a circle, substitute the linear expression for one variable into the circle's equation to get a quadratic equation in the other variable. The discriminant of this quadratic tells you the type of intersection.

**Discriminant Rule for Line-Circle Intersection** — For the resulting quadratic $Ax^2+Bx+C=0$: 1. $\Delta>0$ = two distinct roots (line forms a chord), 2. $\Delta=0$ = one repeated root (line is tangent), 3. $\Delta<0$ = no real roots (line does not meet the circle).

**Worked example:** Determine if the line $y=2x+1$ intersects, is tangent to, or misses the circle $(x-2)^2 + (y-1)^2 = 5$.

1. 1. Substitute $y=2x+1$ into the circle equation:

   $$(x-2)^2 + (2x+1-1)^2 = 5$$
2. 2. Expand and simplify to get a quadratic in x:

   $$x^2 -4x +4 +4x^2 =5 \implies 5x^2 -4x -1 =0$$
3. 3. Calculate the discriminant:

   $$\Delta = (-4)^2 - 4(5)(-1) = 16 +20 =36$$
4. 4. Since $\Delta>0$, the line cuts the circle at two distinct points, forming a chord.

> **Exam tip:** If asked for intersection points, solve the quadratic for x, then substitute back into the line equation to find corresponding y values.

## Equations of Tangents to Circles

> **tip**
>
> For CIE 0606, the expected method is the geometric rule: the radius to the point of tangency is perpendicular to the tangent line. No use of calculus is required.

**Worked example:** Find the equation of the tangent to the circle $(x+1)^2 + (y-3)^2 = 20$ at the point $(1,7)$.

1. 1. Identify the centre of the circle: $(-1, 3)$.
2. 2. Calculate the gradient of the radius connecting the centre to the point of tangency:

   $$m_{radius} = \frac{7-3}{1 - (-1)} = \frac{4}{2}=2$$
3. 3. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal: $m_{tangent} = -\frac{1}{2}$
4. 4. Use point-gradient form of a line with point $(1,7)$:

   $$y -7 = -\frac{1}{2}(x-1)$$
5. 5. Rearrange to standard form: $x + 2y = 15$

> **Exam tip:** Always verify the point of tangency lies on the circle first by substituting it into the circle equation, to avoid mistakes from incorrect points.

## Two Circles: Positional Relationship & Common Chords

For two circles with centres $C_1, C_2$ and radii $r_1, r_2$, the distance between centres $d = |C_1C_2|$ determines their positional relationship:

- $d > r_1 + r_2$: circles are separate, no intersections
- $d = r_1 + r_2$: circles touch externally, 1 common point
- $|r_1 - r_2| < d < r_1 + r_2$: circles intersect at 2 points, have a common chord
- $d = |r_1 - r_2|$: circles touch internally, 1 common point
- $d < |r_1 - r_2|$: one circle is entirely inside the other, no intersections

**Common Chord Equation** — When two circles intersect at two points, the linear equation of their common chord is found by subtracting the expanded general form equations of the two circles, eliminating the $x^2$ and $y^2$ terms.

**Worked example:** Find the equation of the common chord of the two circles $x^2 + y^2 + 2x - 4y -4 = 0$ and $x^2 + y^2 - 2x +6y -10 =0$.

1. 1. Label the circle equations as Equation 1 and Equation 2.
2. 2. Subtract Equation 2 from Equation 1 to eliminate quadratic terms:

   $$(x^2 + y^2 +2x -4y -4) - (x^2 + y^2 -2x +6y -10) = 0$$
3. 3. Simplify the resulting linear equation:

   $$4x -10y +6 =0 \implies 2x -5y +3 =0$$
4. 4. This is the equation of the common chord of the two circles.

> **Exam tip:** Use the common chord equation to find intersection points of two circles by substituting it into either circle's equation and solving the resulting quadratic.

## Common pitfalls

- **Wrong:** Using calculus to find the gradient of a tangent to a circle
  - Why it fails: The 0606 syllabus states that no use of calculus is expected; differentiating a circle equation needs implicit differentiation (beyond 0606), so the geometric method using the perpendicular radius is the reliable route.
  - Correct: Always calculate the gradient of the radius first, then take its negative reciprocal to get the tangent gradient.
- **Wrong:** Reading the centre of the general form circle $x^2+y^2+2gx+2fy+c=0$ as $(g,f)$ instead of $(-g,-f)$
  - Why it fails: Sign errors when reading off centre coordinates from the general form are one of the most common mistakes on this topic.
  - Correct: Write out the values of $2g$ and $2f$ explicitly from the equation, then divide by 2 and flip the sign to get centre coordinates.
- **Wrong:** Calculating the radius from the general form as $\sqrt{g^2 + f^2 + c}$ instead of $\sqrt{g^2 + f^2 - c}$
  - Why it fails: Sign error in the radius formula leads to incorrect radius values, and often non-real radius results if $c$ is negative.
  - Correct: Memorize the radius formula for general form explicitly, and verify it by completing the square if you are unsure.
- **Wrong:** Adding the two circle equations to find the common chord instead of subtracting them
  - Why it fails: Adding does not eliminate the quadratic $x^2$ and $y^2$ terms, so you will not get a linear equation for the common chord.
  - Correct: Always subtract one expanded circle equation from the other to eliminate quadratic terms and get the linear common chord equation.
- **Wrong:** Confusing the conditions for two circles touching internally vs externally
  - Why it fails: Mixing up the sum and difference of radii leads to incorrect classification of the positional relationship of two circles.
  - Correct: Remember that external touch uses the sum of radii, internal touch uses the absolute difference of radii, and check with a quick sketch if possible.

## Cheatsheet

| Concept | Rule/Formula |
| --- | --- |
| Centre-radius circle equation | $(x-a)^2 + (y-b)^2 = r^2$, centre $(a,b)$, radius $r$ |
| General form circle equation | $x^2+y^2+2gx+2fy+c=0$, centre $(-g,-f)$, $r=\sqrt{g^2+f^2-c}$ |
| Line-circle intersection discriminant | $\Delta>0$: chord, $\Delta=0$: tangent, $\Delta<0$: no intersection |
| Tangent gradient rule | $m_{tangent} = -\frac{1}{m_{radius}}$ at point of contact |
| Two circles: external touch | $d = r_1 + r_2$, $d$ = distance between centres |
| Two circles: internal touch | $d = \|r_1 - r_2\|$ |
| Common chord equation | Subtract expanded equations of two intersecting circles |

## What's next

Now that you have mastered circle coordinate geometry for CIE IGCSE Additional Maths 0606, you are ready to apply these skills to complex structured exam problems. This topic is frequently combined with quadratic equations, straight line geometry, and trigonometry in longer exam questions, so revise those prerequisite topics regularly to build speed and accuracy. Practice past paper questions focused on this topic to get familiar with common phrasing and mark scheme requirements, as most questions will ask you to show every step of your working, including discriminant calculations and tangent derivations.

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