Coordinate geometry of the circle

1. Overview

Coordinate geometry of the circle is a crucial topic in IGCSE Additional Mathematics. It combines your algebra skills (especially completing the square) with geometric understanding to analyse circles, lines, and their intersections. Mastering this topic is essential for solving problems involving tangents, chords, and the relationships between multiple circles. Expect to apply these techniques in both Paper 1 (non-calculator) and Paper 2, with an emphasis on exact values and clear algebraic working.

Key Definitions

• Circle: The set of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).
• Radius ($r$): The constant distance from the centre to any point on the circumference.
• Centre ($a, b$): The coordinates of the fixed point in the middle of the circle.
• Tangent: A straight line that touches the circle at exactly one point and is perpendicular to the radius at that point.
• Chord: A straight line segment joining two points on the circumference.
• Discriminant ($b^2 - 4ac$): An algebraic tool used to determine the number of intersection points between a line and a circle.

Core Content

3.1 The Equation of a Circle

There are two main forms of the circle equation.

Standard Form: $$(x - a)^2 + (y - b)^2 = r^2$$

• Centre: $(a, b)$
• Radius: $r$
• Note: This formula is provided on the IGCSE formula sheet.

General Form: $$x^2 + y^2 + 2gx + 2fy + c = 0$$ To find the centre and radius from this form, you must complete the square for both $x$ and $y$ variables.

• Centre: $(-g, -f)$
• Radius: $\sqrt{g^2 + f^2 - c}$

Worked Example 1 — Finding Centre and Radius from General Form

The equation of a circle is given by $x^2 + y^2 - 4x + 6y - 3 = 0$. Find the coordinates of the centre and the exact value of the radius.

Step 1: Group the $x$ and $y$ terms together. $(x^2 - 4x) + (y^2 + 6y) = 3$ Reason: Prepare for completing the square.

Step 2: Complete the square for the $x$ terms. $(x - 2)^2 - 4 + (y^2 + 6y) = 3$ Reason: $(x-2)^2 = x^2 - 4x + 4$, so $(x^2 - 4x) = (x-2)^2 - 4$

Step 3: Complete the square for the $y$ terms. $(x - 2)^2 - 4 + (y + 3)^2 - 9 = 3$ Reason: $(y+3)^2 = y^2 + 6y + 9$, so $(y^2 + 6y) = (y+3)^2 - 9$

Step 4: Rearrange to standard form. $(x - 2)^2 + (y + 3)^2 = 3 + 4 + 9$ $(x - 2)^2 + (y + 3)^2 = 16$ Reason: Isolate the squared terms and constant on each side.

Step 5: Identify the centre and radius. Centre: $(2, -3)$ Radius: $\sqrt{16} = 4$

Final Answer: Centre = $\boxed{(2, -3)}$; Radius = $\boxed{4}$

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3.2 Intersection of a Circle and a Straight Line

To find where a line ($y = mx + c$) meets a circle, substitute the line equation into the circle equation to form a quadratic equation in terms of $x$ (or $y$).

Use the Discriminant ($D = b^2 - 4ac$) of the resulting quadratic to determine the relationship:

1. $D > 0$: Two distinct points of intersection (the line is a chord).
2. $D = 0$: One point of intersection (the line is a tangent).
3. $D < 0$: No points of intersection (the line does not meet the circle).

Worked Example 2 — Finding Intersection Points

The line $y = 2x + 1$ intersects the circle $x^2 + y^2 + 2x - 4y - 5 = 0$. Find the coordinates of the points of intersection.

Step 1: Substitute $y$ from the line equation into the circle equation. $x^2 + (2x + 1)^2 + 2x - 4(2x + 1) - 5 = 0$ Reason: Eliminate $y$ to obtain an equation in $x$ only.

Step 2: Expand and simplify the equation. $x^2 + (4x^2 + 4x + 1) + 2x - 8x - 4 - 5 = 0$ $5x^2 - 2x - 8 = 0$ Reason: Expand the squared term and combine like terms.

Step 3: Solve the quadratic equation for $x$. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{2 \pm \sqrt{(-2)^2 - 4(5)(-8)}}{2(5)}$ $x = \frac{2 \pm \sqrt{4 + 160}}{10}$ $x = \frac{2 \pm \sqrt{164}}{10}$ $x = \frac{2 \pm 2\sqrt{41}}{10}$ $x = \frac{1 \pm \sqrt{41}}{5}$ Reason: The quadratic does not factorise easily, so use the quadratic formula. Simplify the surd.

Step 4: Find the corresponding $y$ values. For $x = \frac{1 + \sqrt{41}}{5}$: $y = 2(\frac{1 + \sqrt{41}}{5}) + 1 = \frac{2 + 2\sqrt{41}}{5} + \frac{5}{5} = \frac{7 + 2\sqrt{41}}{5}$

For $x = \frac{1 - \sqrt{41}}{5}$: $y = 2(\frac{1 - \sqrt{41}}{5}) + 1 = \frac{2 - 2\sqrt{41}}{5} + \frac{5}{5} = \frac{7 - 2\sqrt{41}}{5}$ Reason: Substitute each $x$ value back into the line equation to find the corresponding $y$ value.

Final Answer: The intersection points are $\boxed{\left(\frac{1 + \sqrt{41}}{5}, \frac{7 + 2\sqrt{41}}{5}\right)}$ and $\boxed{\left(\frac{1 - \sqrt{41}}{5}, \frac{7 - 2\sqrt{41}}{5}\right)}$.

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3.3 Tangents to a Circle

A tangent is perpendicular to the radius at the point of contact. You do not need calculus for this.

Method to find the equation of a tangent at point $P(x_1, y_1)$:

1. Find the gradient of the radius ($m_{radius}$) using the centre $(a, b)$ and point $P$.
2. Find the gradient of the tangent ($m_{tan}$) using the perpendicular rule: $m_{tan} = -\frac{1}{m_{radius}}$.
3. Use the point-gradient formula: $y - y_1 = m_{tan}(x - x_1)$.

Worked Example 3 — Equation of a Tangent

A circle has the equation $(x + 1)^2 + (y - 3)^2 = 20$. Find the equation of the tangent to the circle at the point $(1, 7)$. Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers.

Step 1: Identify the centre of the circle. Centre $C = (-1, 3)$. Reason: From the standard form equation $(x - a)^2 + (y - b)^2 = r^2$, the centre is $(a, b)$.

Step 2: Calculate the gradient of the radius $CP$. $m_{radius} = \frac{7 - 3}{1 - (-1)} = \frac{4}{2} = 2$ Reason: Use the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ with points $C(-1, 3)$ and $P(1, 7)$.

Step 3: Determine the gradient of the tangent. $m_{tangent} = -\frac{1}{2}$ Reason: The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of the radius's gradient.

Step 4: Use the point-gradient form to find the equation of the tangent. $y - 7 = -\frac{1}{2}(x - 1)$ Reason: Use the formula $y - y_1 = m(x - x_1)$ with point $P(1, 7)$ and $m_{tangent} = -\frac{1}{2}$.

Step 5: Rearrange the equation into the required form. $2(y - 7) = -(x - 1)$ $2y - 14 = -x + 1$ $x + 2y - 15 = 0$ Reason: Multiply by 2 to eliminate the fraction and rearrange to the form $ax + by + c = 0$.

Final Answer: The equation of the tangent is $\boxed{x + 2y - 15 = 0}$.

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3.4 Intersection of Two Circles

To find the intersection of two circles:

1. Subtract one equation from the other. This eliminates the $x^2$ and $y^2$ terms, leaving a linear equation ($px + qy + r = 0$).
2. This linear equation is the common chord.
3. Substitute this linear equation back into one of the circle equations to find the intersection points.

Determining the relationship between two circles: Let $d$ be the distance between centres and $r_1, r_2$ be the radii.

• Separated: $d > r_1 + r_2$
• Touch externally: $d = r_1 + r_2$
• Intersect at two points: $|r_1 - r_2| < d < r_1 + r_2$
• Touch internally: $d = |r_1 - r_2|$

Worked Example 4 — Intersection of Two Circles

Two circles have equations $x^2 + y^2 - 2x - 4y - 4 = 0$ and $x^2 + y^2 + 4x + 2y - 20 = 0$. Find the equation of their common chord.

Step 1: Subtract the first equation from the second equation. $(x^2 + y^2 + 4x + 2y - 20) - (x^2 + y^2 - 2x - 4y - 4) = 0$ Reason: Subtracting the equations eliminates the $x^2$ and $y^2$ terms.

Step 2: Simplify the resulting equation. $6x + 6y - 16 = 0$ Reason: Combine like terms.

Step 3: Simplify further to find the equation of the common chord. $3x + 3y - 8 = 0$ Reason: Divide the equation by 2 to simplify.

Final Answer: The equation of the common chord is $\boxed{3x + 3y - 8 = 0}$.

Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

Key Equations

Equation	Symbols	Notes
$\mathbf{(x-a)^2 + (y-b)^2 = r^2}$	$(a, b)$ = centre, $r$ = radius	Standard Form (On Formula Sheet)
$\mathbf{x^2 + y^2 + 2gx + 2fy + c = 0}$	Centre $= (-g, -f)$	General Form
$\mathbf{m_1 \cdot m_2 = -1}$	$m$ = gradient	For perpendicular lines (tangents)
$\mathbf{d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}$	$d$ = distance	Distance between two points
$\mathbf{y - y_1 = m(x - x_1)}$	$(x_1, y_1)$ = point, $m$ = gradient	Point-gradient form of a line
$\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$	$a, b, c$ from $ax^2 + bx + c = 0$	Quadratic Formula (On Formula Sheet)

Common Mistakes to Avoid

• ❌ Wrong: Forgetting to square the radius when using the standard form equation. Writing $(x-a)^2 + (y-b)^2 = r$ instead of $(x-a)^2 + (y-b)^2 = r^2$.
• ✓ Right: Always remember that the equation involves $r^2$. If the equation is $(x-1)^2 + (y+2)^2 = 9$, then the radius is $\sqrt{9} = 3$, not $9$.
• ❌ Wrong: Incorrectly identifying the centre from the standard form equation due to sign errors. For example, stating the centre of $(x+5)^2 + (y-2)^2 = 16$ as $(5, 2)$.
• ✓ Right: Remember the standard form is $(x - a)^2 + (y - b)^2 = r^2$. Therefore, $(x + 5)^2$ is equivalent to $(x - (-5))^2$, so $a = -5$. The correct centre is $(-5, 2)$.
• ❌ Wrong: Using the gradient of the radius as the gradient of the tangent.
• ✓ Right: The tangent is perpendicular to the radius. Calculate the gradient of the radius and then take the negative reciprocal to find the gradient of the tangent.
• ❌ Wrong: Giving decimal approximations when the question requires exact values (surds, fractions, or multiples of $\pi$).
• ✓ Right: Leave your answer in surd form (e.g., $\sqrt{7}$), as a simplified fraction (e.g., $\frac{3}{4}$), or in terms of $\pi$ (e.g. $5\pi$) unless the question explicitly asks for a decimal approximation.

Exam Tips

• Exact Values: IGCSE Additional Maths rarely wants decimals. Keep your radii in surd form (e.g., $\sqrt{13}$) unless the question asks for 3 significant figures.
• Command Words: If a question says "Show that the line is a tangent," you must demonstrate that the discriminant $b^2 - 4ac = 0$ after substitution.
• Paper 1 (Non-Calculator): Expect numbers that factorise easily or result in clean surds. Practice completing the square with fractions.
• Check Domain: If finding intersection points, ensure your coordinates satisfy both the line and the circle equations as a final check.
• Completing the Square: Practice completing the square quickly and accurately. This is a fundamental skill for finding the centre and radius from the general form of the circle equation.
• Clear Working: Show all steps in your working, especially in Paper 1. This helps you avoid careless mistakes and allows the examiner to award method marks even if your final answer is incorrect.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The circle $C$ has equation $x^2 + y^2 + 6x - 4y - 12 = 0$.

(a) Find the coordinates of the centre of $C$ and the radius of $C$. [5]

(b) The line $y = x + k$ is a tangent to the circle $C$. Find the possible values of $k$. [3]

Worked Solution:

(a)

1. Complete the square for both $x$ and $y$ terms. $x^2 + 6x + y^2 - 4y = 12$ $(x+3)^2 - 9 + (y-2)^2 - 4 = 12$ $(x+3)^2 + (y-2)^2 = 25$ [Completing the square to get the standard form of the equation]

2. Identify the center and radius from the standard form. Centre: $(-3, 2)$ Radius: $\sqrt{25} = 5$ [Using the standard equation $(x-a)^2 + (y-b)^2 = r^2$ to find the center $(a,b)$ and radius $r$] Centre: $\boxed{(-3, 2)}$ Radius: $\boxed{5}$

How to earn full marks: Show the completing the square steps clearly, and state the center and radius explicitly with correct signs and units.

(b)

1. Substitute $y = x + k$ into the equation of the circle $(x+3)^2 + (y-2)^2 = 25$ $(x+3)^2 + (x+k-2)^2 = 25$ $x^2 + 6x + 9 + x^2 + 2(k-2)x + (k-2)^2 = 25$ $2x^2 + (6 + 2k - 4)x + 9 + k^2 - 4k + 4 - 25 = 0$ $2x^2 + (2k + 2)x + k^2 - 4k - 12 = 0$ [Substituting the line equation into the circle equation]

2. Since the line is a tangent, there is only one solution, so the discriminant must be zero. $(2k+2)^2 - 4(2)(k^2 - 4k - 12) = 0$ $4k^2 + 8k + 4 - 8k^2 + 32k + 96 = 0$ $-4k^2 + 40k + 100 = 0$ $k^2 - 10k - 25 = 0$ $k = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-25)}}{2}$ $k = \frac{10 \pm \sqrt{100 + 100}}{2}$ $k = \frac{10 \pm \sqrt{200}}{2}$ $k = \frac{10 \pm 10\sqrt{2}}{2}$ $k = 5 \pm 5\sqrt{2}$ [Using the discriminant condition for tangency] $k = \boxed{5 + 5\sqrt{2}, 5 - 5\sqrt{2}}$

How to earn full marks: Remember to set the discriminant to zero, show all steps in solving the quadratic, and give both possible values of k in exact form.

Common Pitfall: When finding exact values, avoid using decimals in your working. Keep everything in surd form until the very end. Also, remember to show all steps when using the quadratic formula in a "no calculator" paper.

Exam-Style Question 2 — Paper 2 (Calculator Allowed) [10 marks]

Question:

The points $A(-1, 2)$ and $B(5, 4)$ are the endpoints of a diameter of a circle $C$.

(a) Find the equation of the circle $C$ in the form $x^2 + y^2 + ax + by + c = 0$, where $a$, $b$, and $c$ are constants. [6]

(b) Find the equation of the tangent to the circle at the point $B$. Give your answer in the form $y = mx + c$. [4]

Worked Solution:

(a)

1. Find the midpoint of $AB$, which is the center of the circle. Midpoint $= \left(\frac{-1+5}{2}, \frac{2+4}{2}\right) = (2, 3)$ [Finding the center of the circle using the midpoint formula]

2. Find the radius of the circle. This is the distance from the center to either $A$ or $B$. Radius $= \sqrt{(5-2)^2 + (4-3)^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$ [Calculating the radius using distance formula]

3. Write the equation of the circle in the form $(x-h)^2 + (y-k)^2 = r^2$. $(x-2)^2 + (y-3)^2 = 10$ $x^2 - 4x + 4 + y^2 - 6y + 9 = 10$ $x^2 + y^2 - 4x - 6y + 3 = 0$ [Expanding and rearranging to get required form] $\boxed{x^2 + y^2 - 4x - 6y + 3 = 0}$

How to earn full marks: Show your working for finding the midpoint and radius, and rearrange the equation into the required general form.

(b)

1. Find the gradient of the radius joining the center $(2, 3)$ to the point $B(5, 4)$. Gradient of radius $= \frac{4-3}{5-2} = \frac{1}{3}$ [Finding the gradient of the radius]

2. Find the gradient of the tangent at $B$. The tangent is perpendicular to the radius. Gradient of tangent $= -3$ [Using the property that the tangent is perpendicular to the radius]

3. Use the point-gradient form of a line to find the equation of the tangent. $y - 4 = -3(x - 5)$ $y - 4 = -3x + 15$ $y = -3x + 19$ [Using the point-gradient form to find the tangent equation] $\boxed{y = -3x + 19}$

How to earn full marks: Remember to find the gradient of the radius first, then use the negative reciprocal to find the gradient of the tangent, and express the final answer in the form y = mx + c.

Common Pitfall: Remember that the tangent to a circle is perpendicular to the radius at the point of tangency. Use this fact to find the gradient of the tangent.

Exam-Style Question 3 — Paper 1 (No Calculator Allowed) [9 marks]

Question:

Two circles, $C_1$ and $C_2$, have equations $(x-1)^2 + (y-2)^2 = 4$ and $(x-5)^2 + (y-5)^2 = 1$, respectively.

(a) Find the distance between the centers of the two circles. [2]

(b) Determine whether the two circles intersect, touch, or do not intersect. [3]

(c) Assuming the circles intersect, find the equation of the common chord of the two circles, giving your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers. [4]

Worked Solution:

(a)

1. Identify the centers of the two circles. $C_1$: $(1, 2)$ $C_2$: $(5, 5)$ [Reading the center coordinates directly from the circle equations]

2. Find the distance between the centers. Distance $= \sqrt{(5-1)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ [Using the distance formula] Distance $= \boxed{5}$

How to earn full marks: State the coordinates of both centers clearly, and show the substitution into the distance formula.

(b)

1. Find the radii of the two circles. $r_1 = \sqrt{4} = 2$ $r_2 = \sqrt{1} = 1$ [Reading the radius from the circle equations]

2. Check if the distance between the centers is less than the sum of the radii and greater than the absolute difference of the radii. $r_1 + r_2 = 2 + 1 = 3$ $|r_1 - r_2| = |2 - 1| = 1$ Since the distance between the centres, 5, is greater than the sum of the radii, 3, the circles do not intersect. [Comparing the distance between centers with the sum and difference of the radii] The circles do not intersect.

How to earn full marks: Calculate the sum and difference of the radii, and explicitly compare these values to the distance between the centers to justify your conclusion.

(c) Since the circles do not intersect, there is no common chord.

How to earn full marks: State clearly that since the circles do not intersect, there is no common chord.

Common Pitfall: Always check the condition for intersection before attempting to find the common chord. If the circles don't intersect, there's no common chord to find!

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

Question:

A circle has the equation $(x-3)^2 + (y+1)^2 = 25$. A line $L$ has the equation $y = mx - 3$.

(a) Find the values of $m$ for which the line $L$ is a tangent to the circle. Give your answers to 2 decimal places. [6]

(b) For $m = 0.5$, find the coordinates of the points where the line $L$ intersects the circle. Give your answers to 2 decimal places. [3]

Worked Solution:

(a)

1. Substitute $y = mx - 3$ into the equation of the circle. $(x-3)^2 + (mx - 3 + 1)^2 = 25$ $(x-3)^2 + (mx - 2)^2 = 25$ $x^2 - 6x + 9 + m^2x^2 - 4mx + 4 = 25$ $(1+m^2)x^2 + (-6-4m)x - 12 = 0$ [Substituting the line equation into the circle equation]

2. For the line to be a tangent, the discriminant of the quadratic equation must be zero. $(-6-4m)^2 - 4(1+m^2)(-12) = 0$ $36 + 48m + 16m^2 + 48 + 48m^2 = 0$ $64m^2 + 48m + 84 = 0$ $16m^2 + 12m + 21 = 0$ [Setting the discriminant to zero]

3. Solve the quadratic equation for $m$. $m = \frac{-12 \pm \sqrt{12^2 - 4(16)(21)}}{2(16)}$ $m = \frac{-12 \pm \sqrt{144 - 1344}}{32}$ $m = \frac{-12 \pm \sqrt{-1200}}{32}$ Discriminant is negative, hence there is no real solution.

   Since the discriminant is negative, there is no real solution. [Using quadratic formula to find m]

   There are no values of $m$ for which the line is a tangent.

How to earn full marks: Show the substitution and expansion steps clearly, set up the discriminant equation correctly, and state clearly that there are no real solutions for m because the discriminant is negative.

(b)

1. Substitute $m = 0.5$ into the equation of the line. $y = 0.5x - 3$ [Substituting the value of m]

2. Substitute $y = 0.5x - 3$ into the equation of the circle. $(x-3)^2 + (0.5x - 3 + 1)^2 = 25$ $(x-3)^2 + (0.5x - 2)^2 = 25$ $x^2 - 6x + 9 + 0.25x^2 - 2x + 4 = 25$ $1.25x^2 - 8x - 12 = 0$ [Substituting the line equation into the circle equation]

3. Solve the quadratic equation for $x$. $x = \frac{8 \pm \sqrt{(-8)^2 - 4(1.25)(-12)}}{2(1.25)}$ $x = \frac{8 \pm \sqrt{64 + 60}}{2.5}$ $x = \frac{8 \pm \sqrt{124}}{2.5}$ $x = \frac{8 \pm 11.1355}{2.5}$ $x_1 = \frac{8 + 11.1355}{2.5} = 7.6542 \approx 7.65$ $x_2 = \frac{8 - 11.1355}{2.5} = -1.2542 \approx -1.25$ [Using quadratic formula to find x]

4. Find the corresponding $y$ values. $y_1 = 0.5(7.6542) - 3 = 0.8271 \approx 0.83$ $y_2 = 0.5(-1.2542) - 3 = -3.6271 \approx -3.63$ [Calculating the y values] The points of intersection are $\boxed{(7.65, 0.83)}$ and $\boxed{(-1.25, -3.63)}$.

How to earn full marks: Show all steps in solving the quadratic equation for x, calculate the corresponding y values, and give both coordinates to the specified decimal places.

Common Pitfall: Don't assume there will always be a solution for m in tangency problems. Sometimes, the discriminant is negative, indicating no real values of m exist for the line to be tangent to the circle.