Circle Equations, Lines, Tangents and Two Circles

Q: Do I need to use calculus to find tangents for CIE 0606?

No. For CIE IGCSE Additional Mathematics 0606, you only need to use the geometric rule that the radius to the point of tangency is perpendicular to the tangent line. No differentiation is required or expected in mark schemes.

Q: How do I convert between the two circle equation forms?

To convert from the expanded general form to centre-radius form, complete the square for the and terms separately, then rearrange to match .

Additional Mathematics· 8.1, 8.2, 8.3, 8.4 (2025-2027)· 25 min read

1. Standard Forms of a Circle's Equation★★☆☆☆⏱ 5 min

📘 Definition

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 $(x - 2)^{2} + (y + 3)^{2} = 16$

📘 Definition

Expanded General Form

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

Example:

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

📐 Worked Example

Convert the expanded form equation $x^{2} + y^{2} + 2 x - 4 y - 4 = 0$ 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
1. Complete the square for x terms: $x^{2} + 2 x = (x + 1)^{2} - 1$ , and for y terms: $y^{2} - 4 y = (y - 2)^{2} - 4$
3
1. Substitute back and simplify:
$(x+1)^2 -1 + (y-2)^2 -4 = 4 \implies (x+1)^2 + (y-2)^2 = 9$
4
1. 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.

2. Intersection of a Line and a Circle★★★☆☆⏱ 7 min

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.

📘 Definition

Discriminant Rule for Line-Circle Intersection

For the resulting quadratic $A x^{2} + B x + C = 0$ : 1. $Δ > 0$ = two distinct roots (line forms a chord), 2. $Δ = 0$ = one repeated root (line is tangent), 3. $Δ < 0$ = no real roots (line does not meet the circle).

📐 Worked Example

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

1
1. Substitute $y = 2 x + 1$ into the circle equation:
$(x-2)^2 + (2x+1-1)^2 = 5$
2
1. Expand and simplify to get a quadratic in x:
$x^2 -4x +4 +4x^2 =5 \implies 5x^2 -4x -1 =0$
3
1. Calculate the discriminant:
$\Delta = (-4)^2 - 4(5)(-1) = 16 +20 =36$
4
1. Since $Δ > 0$ , 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.

3. Equations of Tangents to Circles★★★★☆⏱ 7 min

📐 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
1. 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
1. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal: $m_{t an g e n t} = - \frac{1}{2}$
4
1. Use point-gradient form of a line with point $(1, 7)$ :
$y -7 = -\frac{1}{2}(x-1)$
5
1. Rearrange to standard form: $x + 2 y = 15$

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.

4. Two Circles: Positional Relationship & Common Chords★★★☆☆⏱ 6 min

For two circles with centres $C_{1}, C_{2}$ and radii $r_{1}, r_{2}$ , the distance between centres $d = ∣ C_{1} C_{2} ∣$ 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

📘 Definition

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 $x^{2} + y^{2} + 2 x - 4 y - 4 = 0$ and $x^{2} + y^{2} - 2 x + 6 y - 10 = 0$ .

1
1. Label the circle equations as Equation 1 and Equation 2.
2
1. 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
1. Simplify the resulting linear equation:
$4x -10y +6 =0 \implies 2x -5y +3 =0$
4
1. 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.

5. Common Pitfalls

Wrong move:

Using calculus to find the gradient of a tangent to a circle

Why:

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 move:

Always calculate the gradient of the radius first, then take its negative reciprocal to get the tangent gradient.

Wrong move:

Reading the centre of the general form circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ as $(g, f)$ instead of $(- g, - f)$

Why:

Sign errors when reading off centre coordinates from the general form are one of the most common mistakes on this topic.

Correct move:

Write out the values of $2 g$ and $2 f$ explicitly from the equation, then divide by 2 and flip the sign to get centre coordinates.

Wrong move:

Calculating the radius from the general form as $g^{2} + f^{2} + c$ instead of $g^{2} + f^{2} - c$

Why:

Sign error in the radius formula leads to incorrect radius values, and often non-real radius results if $c$ is negative.

Correct move:

Memorize the radius formula for general form explicitly, and verify it by completing the square if you are unsure.

Wrong move:

Adding the two circle equations to find the common chord instead of subtracting them

Why:

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 move:

Always subtract one expanded circle equation from the other to eliminate quadratic terms and get the linear common chord equation.

Wrong move:

Confusing the conditions for two circles touching internally vs externally

Why:

Mixing up the sum and difference of radii leads to incorrect classification of the positional relationship of two circles.

Correct move:

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.

6. Quick Reference 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	$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , centre $(- g, - f)$ , $r = g^{2} + f^{2} - c$
Line-circle intersection discriminant	$Δ > 0$ : chord, $Δ = 0$ : tangent, $Δ < 0$ : no intersection
Tangent gradient rule	$m_{t an g e n t} = - \frac{1}{m _{r a d i u s}}$ 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

7. Frequently Asked

Do I need to use calculus to find tangents for CIE 0606?

No. For CIE IGCSE Additional Mathematics 0606, you only need to use the geometric rule that the radius to the point of tangency is perpendicular to the tangent line. No differentiation is required or expected in mark schemes.

How do I convert between the two circle equation forms?

To convert from the expanded general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ to centre-radius form, complete the square for the $x$ and $y$ terms separately, then rearrange to match $(x - a)^{2} + (y - b)^{2} = r^{2}$ .

Going deeper

syllabusCIE 0606 2025-2027 Official SyllabusRefer to section 8 for circle coordinate geometry requirements

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.