Circles at A-Level
GCSE circle work focused on circumference, area and angle theorems. A-Level combines the same circle theorems with the coordinate geometry you already know — the distance formula (Pythagoras) and gradient rules — to describe a circle as an algebraic equation and solve genuinely new problems.
Four skills make up this topic:
- The equation of a circle — writing and interpreting \((x-a)^2 + (y-b)^2 = r^2\)
- Completing the square — finding the centre and radius from the expanded (general) form
- Circle theorems in coordinates — the angle in a semicircle, and the perpendicular from the centre to a chord
- Tangents — using the fact that a tangent is always perpendicular to the radius at the point of contact
The equation of a circle
A circle with centre \((a, b)\) and radius \(r\) is the set of all points \((x, y)\) a fixed distance \(r\) from the centre. Applying the distance formula (Pythagoras) between \((x, y)\) and \((a, b)\) and squaring both sides gives the standard equation:
\[(x - a)^2 + (y - b)^2 = r^2\]
Worked Example 1
Write down the equation of the circle with centre \((3, -2)\) and radius 5.
1
Substitute \(a = 3\), \(b = -2\), \(r = 5\) into \((x-a)^2 + (y-b)^2 = r^2\)
2
\((x - 3)^2 + (y - (-2))^2 = 5^2\)
Answer\((x-3)^2 + (y+2)^2 = 25\)
Worked Example 2
Find the equation of the circle with centre \(C(1, 2)\) that passes through the point \(P(4, 6)\).
1
The radius is the distance \(CP\): \(r = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
2
Substitute \(a=1\), \(b=2\), \(r=5\) into \((x-a)^2+(y-b)^2=r^2\)
Answer\((x-1)^2 + (y-2)^2 = 25\)
The circle with centre \(C(1, 2)\) and radius 5, passing through \(P(4, 6)\) — the construction found in Worked Example 2.
Completing the square: finding the centre and radius
A circle's equation is often given in expanded (general) form, \(x^2 + y^2 + \text{(terms in }x\text{ and }y\text{)} = 0\), rather than \((x-a)^2+(y-b)^2=r^2\). To find the centre and radius, group the \(x\) terms and \(y\) terms separately and complete the square on each.
Worked Example 3
Find the centre and radius of the circle with equation \(x^2 + y^2 - 6x + 4y - 12 = 0\).
1
Group the \(x\) and \(y\) terms: \((x^2 - 6x) + (y^2 + 4y) = 12\)
2
Complete the square on each: \((x-3)^2 - 9 + (y+2)^2 - 4 = 12\)
3
Rearrange: \((x-3)^2 + (y+2)^2 = 12 + 9 + 4 = 25\)
AnswerCentre \((3, -2)\), radius \(= 5\)
Circle theorems in coordinate geometry
Two GCSE circle theorems reappear constantly in A-Level coordinate geometry, now proved using gradients instead of angle-chasing:
- Angle in a semicircle: if \(AB\) is a diameter of a circle, then for any point \(C\) on the circle, \(\angle ACB = 90°\) — so \(AC\) and \(BC\) have perpendicular gradients.
- Perpendicular from the centre to a chord: the perpendicular from a circle's centre to any chord always bisects that chord.
Worked Example 4
A circle has diameter endpoints \(A(-3, 1)\) and \(B(5, 7)\). Show that the point \(C(5, 1)\) lies on the circle.
1
Gradient of \(AC\): \(A\) and \(C\) share \(y=1\), so \(AC\) is horizontal, gradient \(=0\)
2
Gradient of \(BC\): \(B\) and \(C\) share \(x=5\), so \(BC\) is vertical (undefined gradient)
3
A horizontal line and a vertical line are always perpendicular, so \(\angle ACB = 90°\)
AnswerBy the converse of the angle-in-a-semicircle theorem, \(C\) lies on the circle with diameter \(AB\)
Worked Example 5
A circle has centre \(C(2, 5)\). A chord joins \(A(-2, 1)\) and \(B(6, 1)\). Show that the perpendicular from \(C\) to \(AB\) passes through the midpoint of \(AB\).
1
Midpoint of \(AB\): \(\left(\dfrac{-2+6}{2}, \dfrac{1+1}{2}\right) = (2, 1)\)
2
\(AB\) is horizontal (both points have \(y=1\)), so the perpendicular from \(C\) is vertical: \(x = 2\)
3
The line \(x=2\) passes through \((2, 1)\) — exactly the midpoint found in step 1
AnswerThe perpendicular from \(C\), \(x=2\), passes through the midpoint of \(AB\), confirming the theorem
Tangents to a circle
A tangent touches a circle at exactly one point and is always perpendicular to the radius at that point of contact. To find a tangent's equation: find the gradient of the radius to the given point, take the negative reciprocal, then use \(y - y_1 = m(x-x_1)\).
Worked Example 6
Find the equation of the tangent to the circle \(x^2 + y^2 = 25\) at the point \(P(3, 4)\).
1
The circle has centre \(O(0,0)\). Gradient of radius \(OP\): \(\dfrac{4-0}{3-0} = \dfrac{4}{3}\)
2
The tangent is perpendicular to \(OP\), so its gradient is \(-\dfrac{3}{4}\)
3
Use \(y - y_1 = m(x-x_1)\) at \(P(3,4)\): \(y - 4 = -\dfrac{3}{4}(x - 3)\)
4
Multiply by 4 and rearrange: \(4y - 16 = -3x + 9\)
Answer\(3x + 4y = 25\)
The tangent to \(x^2+y^2=25\) at \(P(3,4)\) meets the radius \(OP\) at a right angle — the construction found in Worked Example 6.
Exam tips
💡 Tip 1
Always check which form the equation is in first
If you see \(x^2 + y^2 + \ldots = 0\) rather than \((x-a)^2+(y-b)^2=r^2\), complete the square before doing anything else — trying to read off the centre from the general form directly is a common error.
💡 Tip 2
Radius ⊥ tangent is the key fact for almost every tangent question
Once you have the gradient of the radius, the tangent gradient is just the negative reciprocal — the rest of the question is identical to any other "find the equation of a line" problem.
💡 Tip 3
Look for horizontal/vertical chords — they make circle theorem proofs much faster
If two points share an \(x\)- or \(y\)-coordinate, the line between them is vertical or horizontal, and perpendicularity to it is immediate — no gradient formula needed.
💡 Tip 4
If the leading coefficients aren't 1, divide through first
An equation like \(2x^2+2y^2-8x+12y-2=0\) must be divided by 2 before completing the square — skipping this step gives the wrong centre and radius.
💡 Tip 5
For line–circle intersection, substitute and count solutions
Substituting the line into the circle's equation gives a quadratic in one variable: two real roots means the line is a chord, one repeated root (discriminant \(=0\)) means it's a tangent, and no real roots means the line misses the circle entirely.
Common mistakes
Common Mistake 1
Reading off the wrong sign for the centre
In \((x-a)^2+(y-b)^2=r^2\), the centre is \((a, b)\) — for \((x+3)^2+(y-5)^2=16\) the centre is \((-3, 5)\), not \((3, 5)\). Always rewrite \(+3\) as \(-(-3)\) mentally before reading off the coordinate.
Common Mistake 2
Forgetting to square root for the radius
The right-hand side of the equation is \(r^2\), not \(r\) — a circle written as \((x-1)^2+(y-2)^2=49\) has radius 7, not 49.
Common Mistake 3
Using the tangent gradient instead of the radius gradient, or vice versa
Always find the radius gradient first, then take the negative reciprocal for the tangent — starting from the wrong one and forgetting to flip it gives a line that's parallel to the tangent, not the tangent itself.
Common Mistake 4
Losing a constant when completing the square on both \(x\) and \(y\)
Each completed square subtracts a different constant (e.g. \(-9\) for the \(x\) terms, \(-4\) for the \(y\) terms) — both must be moved to the other side, not just one.
Common Mistake 5
Assuming a point lies on a circle without checking
"Show that \(C\) lies on the circle" requires an actual check — either substitute the coordinates into the circle's equation, or (for a semicircle) show the angle condition holds. Stating it without justification earns no marks.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Write down the equation of the circle with centre \((4, -1)\) and radius 6.Core Skills
Q2Write down the centre and radius of the circle \((x+2)^2 + (y-5)^2 = 49\).Core Skills
Q3Find the radius of the circle with centre \((0,0)\) that passes through \((6, 8)\).Core Skills
Q4A circle has centre \((3, 4)\) and passes through the origin. Find its radius.Core Skills
Q5Find the centre and radius of the circle \(x^2 + y^2 - 8x + 2y + 8 = 0\).Core Skills
Exam-Style
Q6Find the equation of the circle with centre \(C(-1, 3)\) that passes through the point \((2, 7)\).Exam-Style
Q7Show that the point \((4, 3)\) lies on the circle \(x^2 + y^2 - 4x - 6y + 9 = 0\).Exam-Style
Q8Find the equation of the tangent to the circle \(x^2 + y^2 = 169\) at the point \((5, 12)\).Exam-Style
Q9Find the radius of the circle with equation \(x^2 + y^2 + 6x - 2y - 15 = 0\).Exam-Style
Q10The line \(y = x + 2\) intersects the circle \(x^2 + y^2 = 20\) at two points. Find their coordinates.Exam-Style
A* Challenge
Q11Find the equation of the circle passing through the points \(A(0,0)\), \(B(6,0)\) and \(C(0,8)\).A* Challenge
Q12The line \(y = 2x - 5\) is a tangent to the circle \(x^2 + y^2 = r^2\). Find the value of \(r\), given \(r > 0\).A* Challenge
Q13Find the centre and radius of the circle \(2x^2 + 2y^2 - 8x + 12y - 2 = 0\).A* Challenge
Q14A circle has centre \((5, -2)\) and passes through the point \((1, 1)\). Find the equation of the tangent to the circle at \((1, 1)\).A* Challenge
Q15Two circles have equations \(x^2+y^2=25\) and \((x-8)^2+y^2=9\). Show that the circles touch each other externally.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \((x-4)^2 + (y+1)^2 = 36\)
1
Substitute \(a=4, b=-1, r=6\) into \((x-a)^2+(y-b)^2=r^2\)
\((x-4)^2 + (y+1)^2 = 36\)
Q2 — Centre \((-2, 5)\), radius 7
1
Compare with \((x-a)^2+(y-b)^2=r^2\): \(a=-2\), \(b=5\), \(r^2=49\)
Centre \((-2, 5)\), radius \(=7\)
Q3 — radius \(=10\)
1
\(r = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100}\)
radius \(=10\)
Q4 — radius \(=5\)
1
\(r = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25}\)
radius \(=5\)
Q5 — Centre \((4, -1)\), radius 3
1
\((x^2-8x) + (y^2+2y) = -8\)
2
\((x-4)^2 - 16 + (y+1)^2 - 1 = -8\)
3
\((x-4)^2 + (y+1)^2 = -8+16+1 = 9\)
Centre \((4, -1)\), radius \(=3\)
Exam-Style (Q6–Q10)
Q6 — \((x+1)^2 + (y-3)^2 = 25\)
1
\(r = \sqrt{(2-(-1))^2+(7-3)^2} = \sqrt{9+16} = \sqrt{25} = 5\)
\((x+1)^2 + (y-3)^2 = 25\)
Q7 — Confirmed, \((4,3)\) satisfies the equation
1
Substitute \(x=4, y=3\): \(16 + 9 - 16 - 18 + 9\)
2
\(= 25 - 34 + 9 = 0\)
\((4, 3)\) lies on the circle
Q8 — \(5x + 12y = 169\)
1
Gradient of radius to \((5,12)\): \(\dfrac{12}{5}\), so tangent gradient \(= -\dfrac{5}{12}\)
2
\(y - 12 = -\dfrac{5}{12}(x-5) \implies 12y - 144 = -5x + 25\)
\(5x + 12y = 169\)
Q9 — radius \(=5\)
1
\((x+3)^2 - 9 + (y-1)^2 - 1 = 15\)
2
\((x+3)^2 + (y-1)^2 = 25\)
radius \(=5\)
Q10 — \((-4, -2)\) and \((2, 4)\)
1
Substitute: \(x^2 + (x+2)^2 = 20 \implies 2x^2 + 4x - 16 = 0\)
2
\(x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0\)
3
\(x=-4 \implies y=-2\); \(x=2 \implies y=4\)
\((-4, -2)\) and \((2, 4)\)
A* Challenge (Q11–Q15)
Q11 — \((x-3)^2 + (y-4)^2 = 25\)
1
\(AB\) is along the x-axis, \(AC\) is along the y-axis, so \(\angle A = 90°\)
2
By the converse of the angle-in-a-semicircle theorem, \(BC\) is the diameter
3
Centre \(=\) midpoint of \(BC\) \(= (3, 4)\); radius \(=\) distance from centre to \(A\) \(= \sqrt{9+16}=5\)
\((x-3)^2 + (y-4)^2 = 25\)
Q12 — \(r = \sqrt{5}\)
1
Substitute: \(x^2+(2x-5)^2=r^2 \implies 5x^2 - 20x + (25-r^2) = 0\)
2
Tangency means discriminant \(=0\): \((-20)^2 - 4(5)(25-r^2) = 0\)
3
\(400 - 500 + 20r^2 = 0 \implies 20r^2 = 100\)
\(r = \sqrt{5}\)
Q13 — Centre \((2, -3)\), radius \(\sqrt{14}\)
1
Divide by 2: \(x^2+y^2-4x+6y-1=0\)
2
\((x-2)^2 - 4 + (y+3)^2 - 9 = 1\)
3
\((x-2)^2 + (y+3)^2 = 14\)
Centre \((2, -3)\), radius \(\sqrt{14}\)
Q14 — \(4x - 3y - 1 = 0\)
1
Gradient of radius to \((1,1)\): \(\dfrac{1-(-2)}{1-5} = \dfrac{3}{-4} = -\dfrac{3}{4}\)
2
Tangent gradient \(= \dfrac{4}{3}\) (negative reciprocal)
3
\(y - 1 = \dfrac{4}{3}(x-1) \implies 3y - 3 = 4x - 4\)
\(4x - 3y - 1 = 0\)
Q15 — Confirmed: distance between centres \(=\) sum of radii
1
Centres: \((0,0)\) and \((8,0)\); radii: 5 and 3
2
Distance between centres \(= 8\)
3
Sum of radii \(= 5+3=8\)
Distance \(=\) sum of radii, so the circles touch externally
Summary
- The equation of a circle with centre \((a,b)\) and radius \(r\) is \((x-a)^2+(y-b)^2=r^2\), derived directly from the distance formula.
- If the equation is in general form (\(x^2+y^2+\ldots=0\)), complete the square on the \(x\) and \(y\) terms separately to find the centre and radius — dividing through first if the leading coefficients aren't 1.
- The angle in a semicircle is always \(90°\), and the perpendicular from a circle's centre to a chord always bisects that chord — both provable using gradients.
- A tangent is always perpendicular to the radius at the point of contact: find the radius gradient, take the negative reciprocal, then use \(y-y_1=m(x-x_1)\).
- For a line intersecting a circle, substitute and count real solutions: two roots is a chord, a repeated root (discriminant \(=0\)) is a tangent, no real roots means the line misses the circle.
This topic draws on the completing-the-square technique from Quadratics and the perpendicular gradient condition from Straight Line Graphs. Next in Pure Year 1: Algebraic Methods, which reuses the factorising skills covered here.
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