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The equation of a circle \((x - a)^2 + (y - b)^2 = r^2\), finding the centre and radius by completing the square, and using circle properties: the angle in a semicircle is a right angle, the perpendicular from the centre to a chord bisects the chord, and a tangent is perpendicular to the radius. Circle questions are often long (8 to 10 marks) and combine straight lines, simultaneous equations and the discriminant.
\(x^2 + y^2 + 2gx + 2fy + c = 0\): centre \((-g, -f)\), radius \(\sqrt{g^2 + f^2 - c}\)
Tangent \(\perp\) radius at the point of contact
Perpendicular from the centre bisects a chord
Angle in a semicircle \(= 90^\circ\)
Line and circle: substitute, then \(b^2 - 4ac\): \(\gt 0\) two points, \(= 0\) tangent, \(\lt 0\) none
Point inside the circle if its distance from the centre is less than \(r\)
Notes
The equation of a circle
Centre \((a, b)\), radius \(r\):\((x - a)^2 + (y - b)^2 = r^2\). Centre the origin: \(x^2 + y^2 = r^2\)
Watch the signs: \((x + 4)^2 + (y - 1)^2 = 9\) has centre \((-4, 1)\) and radius 3.
From the expanded form, complete the square for \(x\) and for \(y\): \(x^2 + y^2 + 8x - 2y + 8 = 0\) gives \((x + 4)^2 - 16 + (y - 1)^2 - 1 + 8 = 0\), so \((x + 4)^2 + (y - 1)^2 = 9\)
The right-hand side must be positive for a real circle, which gives conditions on unknown constants: \(x^2 + y^2 - 4x + k = 0\) is \((x - 2)^2 + y^2 = 4 - k\), so \(k \lt 4\)
A point is inside, on or outside the circle if its distance from the centre is less than, equal to or greater than \(r\)
Circle properties
diagram
A tangent at \(P\) is perpendicular to the radius at \(P\)
The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of any chord passes through the centre.
The angle in a semicircle is a right angle: if \(AB\) is a diameter and \(C\) is on the circle, angle \(ACB = 90^\circ\). Conversely, if angle \(ACB = 90^\circ\), then \(AB\) is a diameter of the circle through \(A\), \(B\) and \(C\)
The three circle facts you use most in coordinate geometry.
Lines and circles
diagram
Tangent at \(P\): find the gradient of the radius \(CP\), where \(C\) is the centre, take the negative reciprocal, and use \(y - y_1 = m(x - x_1)\) through \(P\)
Radius gradient 4/3, so the tangent gradient is −3/4 (negative reciprocal).
Intersections: substitute the line into the circle's equation and solve the quadratic.
The discriminant of that quadratic: \(\gt 0\) two points (a chord), \(= 0\) one point (a tangent), \(\lt 0\) no points.
Circle through three points: find where the perpendicular bisectors of two chords meet; that is the centre.
Length of a tangent from an outside point \(T\) to the circle: \(\sqrt{CT^2 - r^2}\), by Pythagoras.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the centre and radius
3 marksGrade C
Group the \(x\) terms and the \(y\) terms.
Complete the square for each and move the constants to the right.
Read off the centre (change the signs) and square root for the radius.
Example. The circle \(C\) has equation \(x^2 + y^2 - 10x + 6y + 9 = 0\). Find the coordinates of the centre of \(C\) and the radius of \(C\)
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\((x - 5)^2 - 25 + (y + 3)^2 - 9 + 9 = 0\)
(M1)
Centre \((5, -3)\)
(A1)
\((x - 5)^2 + (y + 3)^2 = 25\), so radius 5
(A1)
Find the equation of a tangent to a circle
4 marksGrade C
Find the gradient of the radius from the centre to the point.
Take the negative reciprocal for the tangent's gradient.
Use \(y - y_1 = m(x - x_1)\) through the point.
Example. The point \(P(8, 1)\) lies on the circle \((x - 5)^2 + (y + 3)^2 = 25\). Find an equation of the tangent to the circle at \(P\), giving your answer in the form \(ax + by + c = 0\)
Use gradients to show a right angle, or quote the angle in a semicircle.
Example. The points \(A(-1, 2)\) and \(B(7, 8)\) are the ends of a diameter of the circle \(C\). (a) Find an equation for \(C\). (b) The point \(D(0, 9)\) lies on \(C\). Show that angle \(ADB = 90^\circ\)
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(a) Centre \((3, 5)\)
(B1)
\(AB = \sqrt{8^2 + 6^2} = 10\), so \(r = 5\)
(M1)
\((x - 3)^2 + (y - 5)^2 = 25\)
(A1)
(b) Gradient \(AD = 7\) and gradient \(BD = -\dfrac{1}{7}\)
(M1)
\(7 \times \left(-\dfrac{1}{7}\right) = -1\), so \(AD\) is perpendicular to \(BD\) and angle \(ADB = 90^\circ\).
(A1)
Find when a line is a tangent (discriminant)
5 marksGrade A
Substitute the line into the circle's equation.
Collect into a quadratic in \(x\) with coefficients involving the unknown.
Set \(b^2 - 4ac = 0\) and solve.
Example. The line \(y = kx\) is a tangent to the circle \((x - 4)^2 + y^2 = 8\). Find the possible values of \(k\)
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\((x - 4)^2 + k^2x^2 = 8\)
(M1)
\((1 + k^2)x^2 - 8x + 8 = 0\)
(A1)
Tangent: \(b^2 - 4ac = 0\)
(M1)
\(64 - 32(1 + k^2) = 0\)
(A1)
\(k = 1\) or \(k = -1\)
(A1)
Don’t lose marks
Giving \(r^2\) as the radius, e.g. radius 25 instead of 5.
Sign errors in the centre from the completed square.
Forgetting to subtract the squared constants when completing the square.
Using the radius gradient for the tangent instead of its negative reciprocal.
Squaring wrongly: \((kx)^2 = k^2x^2\), not \(kx^2\)
More tips
Memory tricks
The centre's signs flip: \((x + 4)^2\) means the \(x\)-coordinate of the centre is \(-4\)
If you get a negative value for \(r^2\), recheck your completing the square.
Check a tangent: its gradient times the radius gradient should be \(-1\)
Sketch the circle with its centre: it shows which property (tangent, chord or semicircle) to use.
Exam technique
Draw a sketch with the centre, radius and any tangent or chord, and mark right angles.
To show a point lies on a circle, substitute and show the left-hand side equals \(r^2\) exactly, then conclude.
Name the property you use, e.g. 'the tangent is perpendicular to the radius'.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
State the centre and radius from the equation\((x - a)^2 + (y - b)^2 = r^2\) has centre \((a, b)\) and radius \(r\)
Grade D
Write the equation from centre and radiusSubstitute into \((x - a)^2 + (y - b)^2 = r^2\); find \(r\) with the distance formula if needed.
Grade C
Complete the square to find centre and radiuse.g. \(x^2 + y^2 - 6x + 4y - 12 = 0\) becomes \((x - 3)^2 + (y + 2)^2 = 25\): centre \((3, -2)\), radius 5.
Grade C
Find the equation of a tangentThe tangent is perpendicular to the radius at that point, so use the negative reciprocal of the radius gradient.
Grade B
Use chord and semicircle propertiesThe perpendicular bisector of a chord passes through the centre; an angle in a semicircle is \(90^\circ\)
Grade B
Find where a line meets a circleSubstitute the line into the circle's equation and solve the resulting quadratic.
Grade A
Use the discriminant for tangency conditionsThe line is a tangent when the substituted quadratic has \(b^2 - 4ac = 0\), and misses when \(b^2 - 4ac \lt 0\)
Grade A
Find the circle through three pointsThe centre is where the perpendicular bisectors of two chords meet.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The circle \(C\) has equation \(3x^2 + 3y^2 + 36x + 36y + 108 = 0\).
(a) Find the coordinates of the centre of \(C\).[2]
(b) Find the exact radius of \(C\).[2]
Show the answer and mark scheme
(a)Answer: \((-6, -6)\)
M1 for an attempt to complete the square, e.g. \((x + 6)^2 + (y + 6)^2 = \ldots\) (after dividing by 3)
A1 for \((-6, -6)\)
Worked solution: Dividing by 3: \(x^2 + y^2 + 12x + 12y + 36 = 0\) \((x + 6)^2 - 36 + (y + 6)^2 - 36 + 36 = 0\) \((x + 6)^2 + (y + 6)^2 = 36\), so the centre is \((-6, -6)\).
(b)Answer: \(6\)
M1 for \(r^2 = (-6)^2 + (-6)^2 - 36\) (\(= 36\)) using their completed square
A1 for \(r = 6\)
Worked solution: \(r^2 = 36\), so \(r = 6\).
Question 2Medium7 marks
The circle \(C\) has equation \(x^2 + y^2 + 6x + 10y - 7 = 0\).
(a) Find the coordinates of the centre of \(C\).[2]
(b) Find the exact radius of \(C\).[2]
(c) The circle \(C\) meets the \(x\)-axis at the points \(P\) and \(Q\). Find the exact length of \(PQ\).[3]
Show the answer and mark scheme
(a)Answer: \((-3, -5)\)
M1 for an attempt to complete the square, e.g. \((x + 3)^2 + (y + 5)^2 = \ldots\)
A1 for \((-3, -5)\)
Worked solution: \((x + 3)^2 - 9 + (y + 5)^2 - 25 - 7 = 0\) \((x + 3)^2 + (y + 5)^2 = 41\), so the centre is \((-3, -5)\).
(b)Answer: \(\sqrt{41}\)
M1 for \(r^2 = (-3)^2 + (-5)^2 + 7\) (\(= 41\)) using their completed square
A1 for \(r = \sqrt{41}\)
Worked solution: \(r^2 = 41\), so \(r = \sqrt{41}\).
(c)Answer: \(PQ = 8\)
M1 for substituting \(y = 0\) into the equation of \(C\) (or using Pythagoras with the perpendicular from the centre to the chord)
M1 for solving to obtain \(x = -3 \pm 4\) (or half-chord \(= 4\))
A1 for \(8\)
Worked solution: The perpendicular distance from the centre to the \(x\)-axis is \(5\). Half the chord \(= \sqrt{41 - 25} = 4\), so \(PQ = 8\). (Or: \(y = 0\) gives \((x + 3)^2 = 16\), so \(x = -3 \pm 4\).)
Question 3Hard9 marks
For each non-zero real number \(k\), the circle \(C_k\) has centre \((k, 2k)\) and passes through the origin \(O\).
(a) Show that \(C_k\) has equation \(x^2 + y^2 - 2kx - 4ky = 0\).[2]
(b) Prove that all the circles \(C_k\) have the same tangent at \(O\), and find the equation of this tangent.[3]
(c) Find the exact values of \(k\) for which \(C_k\) touches the line \(x = 10\).[4]
Show the answer and mark scheme
(a)Answer: \(r^2 = k^2 + (2k)^2 = 5k^2\); \((x - k)^2 + (y - 2k)^2 = 5k^2\) expands to the given equation.
M1 for the radius from the centre to \(O\): \(r^2 = k^2 + 4k^2 = 5k^2\)
Worked solution: The radius is the distance from \((k, 2k)\) to \(O\): \(r^2 = k^2 + 4k^2 = 5k^2\). \((x - k)^2 + (y - 2k)^2 = 5k^2 \Rightarrow x^2 - 2kx + k^2 + y^2 - 4ky + 4k^2 = 5k^2 \Rightarrow x^2 + y^2 - 2kx - 4ky = 0\).
(b)Answer: The radius to \(O\) has gradient \(\frac{2k}{k} = 2\) for every \(k\), so the tangent at \(O\) has gradient \(-\frac{1}{2}\): \(x + 2y = 0\).
M1 for the gradient of the radius \(O\) to \((k, 2k)\): \(\frac{2k}{k} = 2\), independent of \(k\)
M1 for the tangent being perpendicular to the radius: gradient \(-\frac{1}{2}\)
A1 for \(y = -\frac{1}{2}x\) (or \(x + 2y = 0\)) with a statement that it does not depend on \(k\)
Worked solution: The radius from \(O\) to the centre \((k, 2k)\) has gradient \(\dfrac{2k}{k} = 2\), whatever the value of \(k\). The tangent at \(O\) is perpendicular to this radius, so its gradient is \(-\frac{1}{2}\), and it passes through \(O\): \(y = -\frac{1}{2}x\), i.e. \(x + 2y = 0\), the same for every \(k\).
(c)Answer: \(k = \dfrac{-5 \pm 5\sqrt{5}}{2}\)
M1 for the distance from the centre to the line: \(|10 - k|\)
M1 for equating to the radius: \((10 - k)^2 = 5k^2\)
M1 for solving the quadratic \(k^2 + 5k - 25 = 0\)
A1 for \(k = \frac{-5 \pm 5\sqrt{5}}{2}\)
Worked solution: The line \(x = 10\) is a tangent when its distance from the centre equals the radius: \(|10 - k| = \sqrt{5}|k|\). Squaring: \(100 - 20k + k^2 = 5k^2\), so \(4k^2 + 20k - 100 = 0\), i.e. \(k^2 + 5k - 25 = 0\). \(k = \dfrac{-5 \pm \sqrt{25 + 100}}{2} = \dfrac{-5 \pm 5\sqrt{5}}{2}\).