Coordinate Geometry of Circles, Tangents, Normals & Circle Theorems
Coordinate geometry of circles combines pure algebra (completing the square, simultaneous quadratic equations, and the discriminant b² - 4ac) with Euclidean geometry. Every circle question at A-Level is built around three geometric properties that allow you to avoid messy calculus: first, the tangent to a circle is perpendicular to the radius at the point of contact; second, the angle in a semicircle is a right angle (90°); and third, the perpendicular from the centre of a circle to a chord bisects the chord.
- Standard and Expanded Equations of a Circle
- By Pythagoras' theorem, the locus of all points (x, y) at a constant distance r (radius) from a fixed centre (a, b) satisfies the standard circle equation: (x - a)² + (y - b)² = r². Expanding the brackets gives the general form x² + y² - 2ax - 2by + (a² + b² - r²) = 0, or x² + y² + 2gx + 2fy + c = 0. To find the centre and radius from the expanded form, complete the square separately for the x-terms and the y-terms, and move the constants to the right-hand side to equal r² (requiring r² > 0 for a real circle).
- Equations of Tangents and Normals to a Circle
- You never need differentiation to find the tangent to a circle at a point P(x₁, y₁) on its circumference: 1. Find the gradient of the radius from the centre C(a, b) to P(x₁, y₁): m_radius = (y₁ - b) / (x₁ - a). 2. Because the tangent is perpendicular to the radius, the gradient of the tangent is the negative reciprocal: m_tangent = -1 / m_radius = -(x₁ - a) / (y₁ - b). 3. Substitute P(x₁, y₁) and m_tangent into y - y₁ = m_tangent(x - x₁). Note that the normal to a circle at any point P always passes straight through the centre C(a, b)!
- Intersections of a Line and a Circle (Discriminant Method)
- To find how a straight line y = mx + c interacts with a circle (x - a)² + (y - b)² = r², substitute y = mx + c into the circle equation and expand to obtain a quadratic equation Ax² + Bx + C = 0. Evaluate the discriminant Δ = B² - 4AC: 1. If B² - 4AC > 0, the line intersects the circle at two distinct points (it is a secant/chord); 2. If B² - 4AC = 0, the line touches the circle at one repeated point (it is a tangent); 3. If B² - 4AC < 0, the line does not intersect the circle.
Parallel Line Angles (Z, F, C Shapes) & Polygon Interior Angles
Switch between Alternate (Z), Corresponding (F), Co-interior (C), and Polygon Interior Angles. In the exam, always write the formal geometric name, not the letter!
Alternate Angles (Z-Shape): Equal
Both angles sit on opposite sides of the transversal line between the two parallel lines, forming a "Z" shape. Both equal 62°
- Write “Alternate angles are equal” (never write “Z-angles”).
- Write “Corresponding angles are equal” (never write “F-angles”).
- Write “Co-interior angles add up to 180°” (never write “C-angles”).
Key Concepts & Exam Definitions (6)
- Standard and Expanded Equations of a Circle
- By Pythagoras' theorem, the locus of all points (x, y) at a constant distance r (radius) from a fixed centre (a, b) satisfies the standard circle equation: (x - a)² + (y - b)² = r². Expanding the brackets gives the general form x² + y² - 2ax - 2by + (a² + b² - r²) = 0, or x² + y² + 2gx + 2fy + c = 0. To find the centre and radius from the expanded form, complete the square separately for the x-terms and the y-terms, and move the constants to the right-hand side to equal r² (requiring r² > 0 for a real circle).
- Equations of Tangents and Normals to a Circle
- You never need differentiation to find the tangent to a circle at a point P(x₁, y₁) on its circumference: 1. Find the gradient of the radius from the centre C(a, b) to P(x₁, y₁): m_radius = (y₁ - b) / (x₁ - a). 2. Because the tangent is perpendicular to the radius, the gradient of the tangent is the negative reciprocal: m_tangent = -1 / m_radius = -(x₁ - a) / (y₁ - b). 3. Substitute P(x₁, y₁) and m_tangent into y - y₁ = m_tangent(x - x₁). Note that the normal to a circle at any point P always passes straight through the centre C(a, b)!
- Intersections of a Line and a Circle (Discriminant Method)
- To find how a straight line y = mx + c interacts with a circle (x - a)² + (y - b)² = r², substitute y = mx + c into the circle equation and expand to obtain a quadratic equation Ax² + Bx + C = 0. Evaluate the discriminant Δ = B² - 4AC: 1. If B² - 4AC > 0, the line intersects the circle at two distinct points (it is a secant/chord); 2. If B² - 4AC = 0, the line touches the circle at one repeated point (it is a tangent); 3. If B² - 4AC < 0, the line does not intersect the circle.
- Three essential circle theorems in coordinate geometry
- 1. Angle in a semicircle is 90°: If A, B, and C lie on a circle and ∠ABC = 90° (which you prove by showing m_AB × m_BC = -1 or by satisfying Pythagoras AB² + BC² = AC²), then AC is the diameter of the circle, and the circle's centre is the midpoint of AC. 2. Perpendicular bisector of a chord: The perpendicular bisector of any chord AB always passes through the centre of the circle. 3. Length of a tangent from an external point P(x₀, y₀) to point of contact T: Since ∠CTP = 90°, triangle CTP is right-angled at the tangent point T (not at P!), so PT = √(CP² - r²).
- Finding the Circle Through Three Non-Collinear Points
- Given three points A, B, and C on a circle: first check if the triangle is right-angled (by finding gradients or side lengths squared). If it is right-angled, the hypotenuse is the diameter, making the midpoint the centre and half the hypotenuse the radius. If it is not right-angled, find the equations of the perpendicular bisectors of chord AB and chord BC, and solve them simultaneously: their intersection point is the centre (a, b) of the circumcircle.
- Intersection or Touching of Two Circles
- Let two circles have centres C₁ and C₂ with radii r₁ and r₂ (where r₁ ≥ r₂), and let d = |C₁C₂| be the distance between their centres: 1. The circles touch EXTERNALLY if d = r₁ + r₂; 2. The circles touch INTERNALLY if d = r₁ - r₂; 3. The circles do not intersect (one completely outside the other) if d > r₁ + r₂, or (one completely inside the other) if d < r₁ - r₂; 4. The circles intersect at two distinct points if r₁ - r₂ < d < r₁ + r₂.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
What are the centre and radius of the circle x² + y² - 6x + 10y - 2 = 0?
In right-angled triangle CTP formed by the circle centre C, an external point P, and the point of tangency T, which angle is 90° and which side is the hypotenuse?
If twonormals to a circle at different points P and Q have equations y = 2x + 1 and y = -x + 7, how do you find the centre of the circle?
How do you find the shortest distance from an external point P to any point on a circle of centre C and radius r?
If a circle hascentre (4, -3) and a chord AB has length 16 and radius 10, what is the perpendicular distance d from the centre to the midpoint M of the chord?
What is the condition on k for x² + y² - 4x + 8y + k = 0 to represent a valid circle?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A circle C has equation x² + y² - 8x + 4y - 5 = 0. (a) Find the coordinates of the centre of C and the radius of C. (b) Verify that the point P(7, 2) lies on C and find the equation of the tangent to C at P in the form ax + by + c = 0, where a, b, and c are integers. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Complete the square for x and y: (x - 4)² - 16 + (y + 2)² - 4 - 5 = 0 → (x - 4)² + (y + 2)² = 25 [1 mark]. 2. State the centre (4, -2) and radius r = √25 = 5 [1 mark]. 3. (b) Substitute P(7, 2) into the circle equation: (7 - 4)² + (2 + 2)² = 3² + 4² = 9 + 16 = 25, so P lies on C [1 mark]. 4. Calculate the gradient of the radius from the centre (4, -2) to P(7, 2): m_radius = (2 - (-2)) / (7 - 4) = 4 / 3 [1 mark]. 5. Take the negative reciprocal to find the tangent gradient: m_tangent = -3 / 4 [1 mark]. 6. Substitute P(7, 2) into y - y₁ = m(x - x₁): y - 2 = (-3/4)(x - 7) → 4y - 8 = -3x + 21 → 3x + 4y - 29 = 0 [1 mark].
- Q2
The points A(-2, 1), B(6, 5), and C(4, 9) lie on the circumference of a circle. (a) Show that ∠ABC = 90°. (b) Hence find the equation of the circle passing through A, B, and C. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Calculate the gradient of AB and the gradient of BC: m_AB = (5 - 1) / (6 - (-2)) = 4 / 8 = 1/2, and m_BC = (9 - 5) / (4 - 6) = 4 / (-2) = -2 [1 mark]. 2. Multiply the gradients: m_AB × m_BC = (1/2) × (-2) = -1; therefore AB is perpendicular to BC and ∠ABC = 90° [1 mark]. 3. (b) Since the angle in a semicircle is 90°, the hypotenuse AC is a diameter of the circle [1 mark]. 4. Find the centre of the circle as the midpoint M of AC: M = ((-2 + 4)/2, (1 + 9)/2) = (1, 5) [1 mark]. 5. Find r² as the squared distance from the centre M(1, 5) to A(-2, 1): r² = (-2 - 1)² + (1 - 5)² = 9 + 16 = 25; hence the circle equation is (x - 1)² + (y - 5)² = 25 [1 mark].
- Q3
The line y = 2x + k is a tangent to the circle (x - 3)² + (y + 1)² = 20. Find the two possible exact values of the constant k. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Substitute y = 2x + k into the circle equation: (x - 3)² + (2x + k + 1)² = 20 [1 mark]. 2. Expand the brackets and collect terms into a quadratic in x: (x² - 6x + 9) + [4x² + 4(k + 1)x + (k + 1)²] - 20 = 0 → 5x² + (4k - 2)x + (k² + 2k - 10) = 0 [1 mark]. 3. Since the line is a tangent to the circle, set the discriminant b² - 4ac = 0: (4k - 2)² - 4(5)(k² + 2k - 10) = 0 [1 mark]. 4. Expand and simplify the quadratic in k: (16k² - 16k + 4) - (20k² + 40k - 200) = 0 → -4k² - 56k + 204 = 0 → k² + 14k - 51 = 0 [1 mark]. 5. Factorise to find k: (k + 17)(k - 3) = 0 → k = -17 or k = 3 [1 mark].
- Q4
From the external point P(10, 9), two tangents PQ and PR are drawn to the circle C with equation (x - 2)² + (y - 3)² = 36, touching the circle at Q and R. Find the exact length of each tangent PQ, and calculate the area of the kite PQCR, where C is the centre of the circle. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Identify the centre C(2, 3) and radius r = √36 = 6 from the circle equation [1 mark]. 2. Calculate the distance CP between C(2, 3) and P(10, 9): CP = √[(10 - 2)² + (9 - 3)²] = √(64 + 36) = √100 = 10 [1 mark]. 3. Since the radius CQ is perpendicular to the tangent PQ at Q (∠CQP = 90°), apply Pythagoras' theorem with CP as the hypotenuse: PQ = √(CP² - r²) = √(10² - 6²) = √64 = 8 [1 mark]. 4. The kite PQCR consists of two congruent right-angled triangles (△CQP and △CRP), each of area (1/2) × base × height = (1/2) × 6 × 8 = 24; therefore Total Area of kite PQCR = 2 × 24 = 48 square units [1 mark].
Common Exam Questions (4)
What is the biggest mistake students make when using Pythagoras on a tangent from an external point P?
Treating the tangent PT as the hypotenuse instead of the line segment CP joining the centre to the external point. Because the right angle is always at the point of contact T between the radius and the tangent, CP is opposite the 90° angle and is always the hypotenuse: PT² = CP² - r².
What if the radius is vertical or horizontal when I try to find the tangent equation?
If the point of contact P lies directly above or below the centre (e.g., centre (3, 2) and P(3, 7)), the radius is a vertical line (x = 3), so the tangent is simply the horizontal line y = 7! Similarly, if P is directly to the left or right of the centre (e.g., P(8, 2)), the tangent is the vertical line x = 8.
How do I find the common chord (or points of intersection) of two intersecting circles?
Write both circle equations in expanded form (x² + y² + .) and subtract one equation from the other. The x² and y² terms immediately cancel out, leaving the linear equation of the common chord, which you can then rearrange for y and substitute back into one of the circles.
How do I find the range of values of a parameter k for which a line intersects a circle at two distinct points?
Substitute the line equation into the circle equation, expand into a quadratic Ax² + Bx + C = 0, and solve the quadratic inequality B² - 4AC > 0.
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