Intersection Problems

This lesson covers finding intersection points between lines and curves, between two curves, and using parametric equations in intersection problems. These techniques are essential for the Edexcel 9MA0 specification and combine algebra with coordinate geometry.

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Line Meets Curve

To find the intersection of a line and a curve, substitute the equation of the line into the equation of the curve (or vice versa) and solve the resulting equation.

Example: Line and Parabola

Find where y = 2x − 1 meets y = x² − 2x + 3.

Set equal: 2x − 1 = x² − 2x + 3 0 = x² − 4x + 4 0 = (x − 2)² x = 2 (repeated root)

When x = 2: y = 3. The line is tangent to the parabola at (2, 3).

Example: Line and Circle

Find where 2x + y = 10 meets x² + y² = 50.

From the line: y = 10 − 2x. Substitute: x² + (10 − 2x)² = 50 x² + 100 − 40x + 4x² = 50 5x² − 40x + 50 = 0 x² − 8x + 10 = 0

x = (8 ± √(64 − 40))/2 = (8 ± √24)/2 = 4 ± √6.

The two x-values give two intersection points.

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Curve Meets Curve

When two curves intersect, set their y-values (or x-values) equal and solve.

Example: Find the intersection of y = x² and y = 4x − x².

Set equal: x² = 4x − x² 2x² − 4x = 0 2x(x − 2) = 0 x = 0 or x = 2

Points: (0, 0) and (2, 4).

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Intersection Using Parametric Equations

When a parametric curve intersects a Cartesian curve, substitute the parametric expressions into the Cartesian equation.

Example: The curve x = 2t, y = t² meets the line y = 3x − 4. Find the points of intersection.

Substitute: t² = 3(2t) − 4 t² = 6t − 4 t² − 6t + 4 = 0 t = (6 ± √(36 − 16))/2 = (6 ± √20)/2 = 3 ± √5

For t = 3 + √5: x = 6 + 2√5, y = (3 + √5)² = 14 + 6√5. For t = 3 − √5: x = 6 − 2√5, y = (3 − √5)² = 14 − 6√5.

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Using the Discriminant

The discriminant of the resulting equation tells you the nature of the intersection:

Discriminant	Meaning
b² − 4ac > 0	Two intersection points
b² − 4ac = 0	One point (tangency)
b² − 4ac < 0	No intersection

This is especially powerful when the question involves an unknown parameter (like k).

Example: For what value of k is y = kx tangent to y = x² + 2x + 5?

Set equal: kx = x² + 2x + 5 x² + (2 − k)x + 5 = 0

For tangency: (2 − k)² − 20 = 0 4 − 4k + k² = 20 k² − 4k − 16 = 0 k = (4 ± √(16 + 64))/2 = (4 ± √80)/2 = 2 ± 2√5

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Finding the Chord Length

If a line intersects a curve at two points, the distance between them can be found using the distance formula.

If the two x-coordinates are x₁ and x₂ and the line has gradient m, then the chord length is:

Length = √(1 + m²) × |x₁ − x₂|

Example: A line with gradient 3 intersects a curve at x = 1 and x = 4. The chord length is:

|x₁ − x₂| = 3, so length = √(1 + 9) × 3 = 3√10.

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Simultaneous Equations in Coordinate Geometry

Many coordinate geometry problems reduce to simultaneous equations:

• Finding where a tangent from an external point touches a circle.
• Finding the common chord of two circles.
• Determining the intersection of a normal with the original curve.

Example: Find the common chord of x² + y² = 25 and x² + y² − 6x − 2y = 0.

Subtract the second from the first: 6x + 2y − 25 = 0. This is the equation of the common chord.

Exam Tip: The common chord of two circles is found by subtracting one circle equation from the other. The result is always a straight line (the radical axis).

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Key Terms

Term	Definition
Intersection point	A point that lies on two or more curves simultaneously
Tangent (to a curve)	A line that touches a curve at exactly one point
Secant	A line that crosses a curve at two points
Chord	The line segment between two intersection points on a curve
Radical axis	The locus of points with equal power with respect to two circles

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Summary

• Find intersections by substituting one equation into another and solving.
• The discriminant determines whether there are 0, 1 or 2 intersections.
• For parametric curves, substitute the parametric expressions into the other equation.
• Use discriminant = 0 to find tangency conditions with unknown parameters.
• The common chord of two circles is found by subtracting their equations.

A-Level Deep Dive: Intersection Problems

Spec mapping

Edexcel 9MA0-01 specification section 4 — Coordinate geometry in the (x, y) plane, sub-strands 4.1 and 4.2 covers the equation of a straight line; understand and use the coordinate geometry of the circle including using the equation of a circle in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 (refer to the official specification document for exact wording). The intersection of geometric loci sits at the structural heart of A-Level Pure Mathematics — it is where section 4 (Coordinate geometry) meets section 2 (Algebra and functions, sub-strand 2.3 — solving simultaneous equations) through substitution and elimination, where it meets section 2.4 (the quadratic discriminant $b^2 - 4ac$b2−4ac) to classify intersections as two-point (secant), one-point (tangent) or none, and where it meets section 8 (Parametric curves) when curves are given in parametric form. Calculus-based tangent conditions appear in section 9 (Differentiation) when the gradient at a point of contact must equal the gradient of a candidate tangent line. The Edexcel formula booklet provides the quadratic formula but does not list the discriminant interpretation rules — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

The line $l$l has equation $y = 2x - 1$y=2x−1. The circle $C$C has equation $(x - 3)^2 + (y - 2)^2 = 14$(x−3)2+(y−2)2=14.

(a) Show that the x-coordinates of any points of intersection of $l$l and $C$C satisfy $5x^2 - 18x + 4 = 0$5x2−18x+4=0. (4)

(b) Hence find the coordinates of the points of intersection of $l$l and $C$C, giving each coordinate as an exact value. (4)

Solution with mark scheme:

(a) Step 1 — substitute the line equation into the circle equation.

Replace $y$y in the circle equation by $2x - 1$2x−1:

$(x - 3)^2 + ((2x - 1) - 2)^2 = 14$(x−3)2+((2x−1)−2)2=14 $(x - 3)^2 + (2x - 3)^2 = 14$(x−3)2+(2x−3)2=14

M1 — substituting the linear expression for $y$y into the circle equation. Examiners look for one equation in $x$x alone after this step. A common error is to substitute only the $y^2$y2 term, leaving a hybrid equation in both $x$x and $y$y.

Step 2 — expand both squared brackets.

$(x - 3)^2 = x^2 - 6x + 9$(x−3)2=x2−6x+9.

$(2x - 3)^2 = 4x^2 - 12x + 9$(2x−3)2=4x2−12x+9.

M1 — correct expansion of both brackets. The cross-term in $(2x - 3)^2$(2x−3)2 is $2 \cdot 2x \cdot (-3) = -12x$2⋅2x⋅(−3)=−12x, not $-6x$−6x — a frequent slip.

Step 3 — collect and rearrange.

$x^2 - 6x + 9 + 4x^2 - 12x + 9 = 14$x2−6x+9+4x2−12x+9=14 $5x^2 - 18x + 18 = 14$5x2−18x+18=14 $5x^2 - 18x + 4 = 0$5x2−18x+4=0

M1 — collecting like terms cleanly to standard quadratic form.

A1 — final form matching the printed quadratic.

(b) Step 1 — solve the quadratic.

Apply the quadratic formula to $5x^2 - 18x + 4 = 0$5x2−18x+4=0:

$x = \dfrac{18 \pm \sqrt{324 - 80}}{10} = \dfrac{18 \pm \sqrt{244}}{10} = \dfrac{18 \pm 2\sqrt{61}}{10} = \dfrac{9 \pm \sqrt{61}}{5}$x=1018±324−80​​=1018±244​​=1018±261​​=59±61​​

M1 — correct application of the quadratic formula. Discriminant $324 - 80 = 244 > 0$324−80=244>0, confirming two real intersection points (a secant line).

A1 — correct simplified surd form, with $\sqrt{244} = 2\sqrt{61}$244​=261​ extracted (since $244 = 4 \cdot 61$244=4⋅61).

Step 2 — find corresponding y-coordinates from the line equation.

For $x = \dfrac{9 + \sqrt{61}}{5}$x=59+61​​: $y = 2x - 1 = \dfrac{18 + 2\sqrt{61}}{5} - 1 = \dfrac{13 + 2\sqrt{61}}{5}$y=2x−1=518+261​​−1=513+261​​

For $x = \dfrac{9 - \sqrt{61}}{5}$x=59−61​​: $y = \dfrac{13 - 2\sqrt{61}}{5}$y=513−261​​

M1 — substituting back into the line equation (cheaper than the circle, since linear).

A1 — both points stated as ordered pairs: $\left(\dfrac{9 + \sqrt{61}}{5}, \dfrac{13 + 2\sqrt{61}}{5}\right), \quad \left(\dfrac{9 - \sqrt{61}}{5}, \dfrac{13 - 2\sqrt{61}}{5}\right)$(59+61​​,513+261​​),(59−61​​,513−261​​)

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = x^2 - 4x + 7$y=x2−4x+7 and the line $l$l has equation $y = 2x + k$y=2x+k, where $k$k is a real constant.

(a) Show that the x-coordinates of any points of intersection satisfy $x^2 - 6x + (7 - k) = 0$x2−6x+(7−k)=0. (2)

(b) Find the value of $k$k for which $l$l is a tangent to $C$C, and find the coordinates of the point of contact. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — equating $x^2 - 4x + 7 = 2x + k$x2−4x+7=2x+k and rearranging.
• A1 (AO1.1b) — correct printed form $x^2 - 6x + (7 - k) = 0$x2−6x+(7−k)=0.

(b)

• M1 (AO2.1) — recognising tangency requires the discriminant of the simultaneous quadratic to be zero: $b^2 - 4ac = 0$b2−4ac=0.
• M1 (AO1.1b) — substituting $a = 1$a=1, $b = -6$b=−6, $c = 7 - k$c=7−k: $36 - 4(7 - k) = 0$36−4(7−k)=0, giving $36 - 28 + 4k = 0$36−28+4k=0, so $4k = -8$4k=−8, hence $k = -2$k=−2.
• A1 (AO1.1b) — for tangency $x = -b/(2a) = 6/2 = 3$x=−b/(2a)=6/2=3 (repeated root); then $y = 2(3) + (-2) = 4$y=2(3)+(−2)=4.
• A1 (AO2.5) — point of contact $(3, 4)$(3,4) stated explicitly.

Total: 6 marks split AO1 = 4, AO2 = 2. This is a structurally classic Paper 1 problem: the tangent condition is examined through the discriminant, not through calculus, and the AO2 marks are reserved for recognising that "tangent" translates to "discriminant equals zero".

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Synoptic links

Connects to:

1. Section 2 — Algebra and functions (simultaneous equations): every intersection problem reduces to solving simultaneous equations by substitution. The line-circle case substitutes a linear expression for $y$y into a quadratic-in-$x$x-and-$y$y relation. The line-parabola case is identical structurally. Comfort with eliminating one variable cleanly is the prerequisite.

2. Section 2.4 — The quadratic discriminant $b^2 - 4ac$b2−4ac: the entire taxonomy of intersection counts (two points, one point, no points) is encoded in the sign of the discriminant of the resulting quadratic. $\Delta > 0$Δ>0 — secant; $\Delta = 0$Δ=0 — tangent; $\Delta < 0$Δ<0 — no real intersection. This is one of the most-tested synoptic ideas in 9MA0-01.

3. Section 4.2 — Circles: the circle equation $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2 is what produces the quadratic when intersected with a line. Knowing how to complete the square to recover this form from a general expansion $x^2 + y^2 + Dx + Ey + F = 0$x2+y2+Dx+Ey+F=0 is essential — many exam questions present circles in the expanded form.

4. Section 8 — Parametric curves: intersections involving parametric curves require eliminating the parameter or substituting the parametric expressions into the Cartesian equation of the second curve. The same discriminant logic applies once a single-variable equation in the parameter $t$t has been formed.

5. Section 9 — Differentiation (tangent condition): an alternative route to finding tangents is to differentiate the curve, set the derivative equal to the gradient of the candidate tangent line, and solve for the point of contact. The discriminant route and the calculus route should give identical answers — checking both is an A* habit.

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Mark-scheme literacy

Intersection-problem questions on 9MA0 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–65%	Substituting cleanly, expanding brackets correctly, applying the quadratic formula, computing discriminants
AO2 (reasoning / interpretation)	25–40%	Identifying which simultaneous-equation method is cheapest, translating "tangent" into "discriminant zero", interpreting the sign of $\Delta$Δ as a geometric statement
AO3 (problem-solving)	5–15%	Open-ended modelling — typically when intersections of curves arise inside a longer mechanics or pure problem

Examiner-rewarded phrasing: "since the discriminant $b^2 - 4ac = 0$b2−4ac=0, the line is tangent to the curve at the repeated root"; "substituting $y = mx + c$y=mx+c into the circle equation yields a quadratic in $x$x whose discriminant determines the number of intersections"; "the simpler equation (the line) is used to recover $y$y once $x$x is known". Phrases that lose marks: writing "no solutions" when the discriminant is negative without stating "no real solutions" (complex roots exist but are not geometric intersections in the real plane); equating gradients without checking the point lies on both curves; abandoning surd form for decimal approximations when an exact answer is requested.

A specific Edexcel pattern to watch: questions asking "find the values of $k$k for which the line is tangent to the curve" demand the set of discriminant-zero solutions — usually two values when $k$k enters the constant term linearly. Stopping at one value loses an A1.

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Grade-band model answers

3-mark question

Question: Find the coordinates of the points where the line $y = x + 2$y=x+2 meets the curve $y = x^2$y=x2.

Grade C response (~210 words):

Set the equations equal: $x^2 = x + 2$x2=x+2.

Rearrange: $x^2 - x - 2 = 0$x2−x−2=0.

Factorise: $(x - 2)(x + 1) = 0$(x−2)(x+1)=0.

So $x = 2$x=2 or $x = -1$x=−1.

When $x = 2$x=2: $y = 2 + 2 = 4$y=2+2=4, giving the point $(2, 4)$(2,4).

When $x = -1$x=−1: $y = -1 + 2 = 1$y=−1+2=1, giving the point $(-1, 1)$(−1,1).

So the line meets the curve at $(2, 4)$(2,4) and $(-1, 1)$(−1,1).