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Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Tangents, chords and intersections
Pure content (sections A–J) · Subtopic 9 of 66

Tangents, Chords and Intersections

The two right angles that unlock circle questions (tangent and radius, centre and chord), and how substituting a line into a circle or curve turns geometry into a quadratic whose discriminant tells you how they meet.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Find the equation of the tangent and the normal to a circle at a given point.
  • Use the perpendicular from the centre to bisect a chord, and find chord and tangent lengths.
  • Find where a line meets a circle or a curve by substitution.
  • Use the discriminant to decide whether a line is a tangent, a chord line, or misses completely.
Pause and think

A bicycle wheel rests on flat ground. Where is the spoke that points at the contact point, and what angle does it make with the ground?

Section 2 of 14

Core idea

Start hereDoes the road clip the roundabout?

A straight road passes near a circular roundabout. Does it cut through, just graze the edge, or miss entirely? Substitute the road's equation into the circle's and you get a quadratic. Its discriminant gives the answer: two crossings, one (a tangent), or none.

The big idea

Substitute the line into the circle's equation to get a quadratic. Its roots give the intersection points; its discriminant says whether the line cuts (Δ>0\Delta>0), touches (Δ=0\Delta=0) or misses (Δ<0\Delta<0). A tangent is perpendicular to the radius at the point of contact.

1. Intersections

Line y = k crossing the circle x² + y² = 9
−6−4−2246−4−2240xy
Substituting y=ky=k gives x2=9−k2x^2=9-k^2: two points when ∣k∣<3|k|<3, one (tangent) when ∣k∣=3|k|=3, none when ∣k∣>3|k|>3.
Worked exampley=x+1y=x+1 and x2+y2=13x^2+y^2=13

Find where the line meets the circle.

Goal
All intersection points.
Strategy
Substitute, solve, back-substitute into the line.
  1. x2+(x+1)2=13x^2+(x+1)^2=13
    Replace yy.
  2. 2x2+2x+1=13⇒x2+x−6=02x^2+2x+1=13\Rightarrow x^2+x-6=0
    Expand, rearrange, divide by 22.
  3. (x+3)(x−2)=0(x+3)(x-2)=0
    Factorise.
  4. x=2⇒y=3;x=−3⇒y=−2x=2\Rightarrow y=3;\quad x=-3\Rightarrow y=-2
    Use the line.
Answer(2,3)(2,3) and (−3,−2)(-3,-2).
✓ Check4+9=134+9=13 ✓; 9+4=139+4=13 ✓.

2. Tangency condition

Worked exampleFor which kk is y=x+ky=x+k a tangent to x2+y2=8x^2+y^2=8?

Find kk.

Goal
kk giving exactly one intersection.
Strategy
Substitute; set Δ=0\Delta=0.
  1. x2+(x+k)2=8⇒2x2+2kx+k2−8=0x^2+(x+k)^2=8\Rightarrow2x^2+2kx+k^2-8=0
    Substitute and expand.
  2. (2k)2−4(2)(k2−8)=0(2k)^2-4(2)(k^2-8)=0
    Δ=0\Delta=0.
  3. 4k2−8k2+64=0⇒k2=16⇒k=±44k^2-8k^2+64=0\Rightarrow k^2=16\Rightarrow k=\pm4
    Simplify.
Answerk=4k=4 or k=−4k=-4.
✓ CheckDistance from the centre (0,0)(0,0) to x−y+4=0x-y+4=0 is 42=22=\frac{4}{\sqrt2}=2\sqrt2= radius ✓.

3. The tangent at a point

A tangent meets the radius at 90°
−10−5510−6−4−224680xytangent90°centreP(3, 4)radius
Radius gradient 43\tfrac43; tangent gradient −34-\tfrac34. Their product is −1-1. The tangent at PP is y−4=−34(x−3)y-4=-\tfrac34(x-3).
Why a tangent is perpendicular to the radius
  1. Let T touch the circle at P, centre C.\text{Let }T\text{ touch the circle at }P\text{, centre }C.
    Setup.
  2. Every other point Q of the tangent is outside: CQ>r=CP\text{Every other point }Q\text{ of the tangent is outside: }CQ>r=CP
    A tangent meets the circle only at PP.
  3. CP is the shortest distance from C to the lineCP\text{ is the shortest distance from }C\text{ to the line}
    It's shorter than every other CQCQ.
  4. CP⊥tangentCP\perp\text{tangent}
    The shortest distance to a line is along the perpendicular.
Worked exampleTangent to (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25 at P(4,2)P(4,2)

Find the tangent's equation.

Goal
A line through PP perpendicular to CPCP.
Strategy
Radius gradient → negative reciprocal → point–gradient.
  1. 9+16=25 ✓9+16=25\ \checkmark
    PP is on the circle.
  2. mCP=2−(−2)4−1=43m_{CP}=\frac{2-(-2)}{4-1}=\frac43
    Centre C(1,−2)C(1,-2).
  3. m=−34m=-\tfrac34
    Perpendicular.
  4. y−2=−34(x−4)⇒3x+4y−20=0y-2=-\tfrac34(x-4)\Rightarrow3x+4y-20=0
    Multiply by 44, rearrange.
Answer3x+4y−20=03x+4y-20=0.
✓ CheckPP: 12+8−20=012+8-20=0 ✓.
Worked exampleLength of a tangent from an outside point

Find the length of the tangent from T(7,1)T(7,1) to x2+y2=25x^2+y^2=25.

Goal
Distance from TT to the contact point.
Strategy
The radius, tangent and CTCT form a right-angled triangle.
  1. CT2=72+12=50CT^2=7^2+1^2=50
    Centre at the origin.
  2. tangent2=CT2−r2=50−25=25\text{tangent}^2=CT^2-r^2=50-25=25
    Pythagoras, right angle at the contact point.
Answer55.
✓ CheckBoth tangents from TT have this same length (symmetry).
Pause and think

Does y=2x+10y=2x+10 meet x2+y2=4x^2+y^2=4?

Summary

  • Substitute → quadratic → discriminant.
  • Tangent ⟂ radius.
  • Tangent length d2−r2\sqrt{d^2-r^2}.
Key words
Tangent
A line touching the circle at one point.
Normal
The line through the contact point and the centre.
HistoryTangent means "touching". The word comes from the Latin tangere, "to touch". Euclid proved around 300 BC that the line at right angles to a diameter, through its end, touches the circle and never enters it. Two thousand years later, finding tangents to any curve was the problem that led Newton and Leibniz to calculus.
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