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Circle propertiesEdexcel International A Level Maths: Revision notes

Section 1

Three properties you may use

Edexcel IAL expects you to use three circle properties together with coordinate geometry (gradients, midpoints, distances and equations of lines):

  • the angle in a semicircle is a right angle;
  • the perpendicular from the centre to a chord bisects the chord;
  • a tangent is perpendicular to the radius at the point of contact. Each one turns a geometric fact into an algebraic condition: two lines are perpendicular when the product of their gradients is −1-1, i.e. m1m2=−1m_1m_2=-1. The circle equation is (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2, with centre (a,b)(a,b) and radius rr.
Key termssemicirclechordtangentradius

Section 2

The angle in a semicircle

If ABAB is a diameter and PP is on the circle, then angle APB=90∘APB=90^\circ. In coordinates, find the gradients of APAP and BPBP; if their product is −1-1 the lines are perpendicular. For A=(1,2)A=(1,2), B=(9,8)B=(9,8), P=(8,1)P=(8,1): mAP=−17m_{AP}=-\frac17 and mBP=7m_{BP}=7, product −1-1, so PP is on the circle with diameter ABAB. The converse also works: if angle APB=90∘APB=90^\circ, then PP lies on the circle with diameter ABAB, whose centre is the midpoint of ABAB and whose radius is 12AB\frac12AB.

Key termsdiameterperpendicular
Exam tip

To find a circle from the ends of a diameter, use the midpoint for the centre and half the distance ABAB for the radius.

Section 3

Chords and the perpendicular from the centre

The perpendicular from the centre CC to a chord PQPQ meets PQPQ at its midpoint MM. This gives two routes:

  • the line CMCM has gradient −1mPQ-\frac{1}{m_{PQ}} and passes through CC, so you can find MM without solving for PP and QQ;
  • triangle CMQCMQ is right-angled at MM, so CM2+MQ2=r2CM^2+MQ^2=r^2. Worked example: circle centre (2,3)(2,3), radius 55, chord PQPQ with P=(5,7)P=(5,7), Q=(6,0)Q=(6,0). PQ=50PQ=\sqrt{50}, so MQ2=12.5MQ^2=12.5 and CM2=25−12.5=12.5CM^2=25-12.5=12.5, giving CM=522CM=\frac{5\sqrt2}{2}.
Key termsmidpointbisects
Common mistake

Using the radius as the distance from the centre to the chord. The distance to the chord is shorter than rr.

Section 4

Tangent and radius

The tangent at a point TT on a circle is perpendicular to the radius CTCT. To find the tangent at TT:

  1. find the gradient mm of the radius CTCT;
  2. the tangent has gradient −1m-\frac1m;
  3. use y−y1=m(x−x1)y-y_1=m(x-x_1) with the point TT. Example: (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25 at T=(4,2)T=(4,2). The radius gradient is 2−(−2)4−1=43\frac{2-(-2)}{4-1}=\frac43, so the tangent gradient is −34-\frac34 and y−2=−34(x−4)y-2=-\frac34(x-4), i.e. 3x+4y−20=03x+4y-20=0. If the radius is horizontal the tangent is vertical (x=kx=k), and vice versa.
Key termspoint of contact
Common mistake

Using the gradient of the radius as the gradient of the tangent. You must take the negative reciprocal.

Section 5

Putting it together in exam questions

Typical problems ask you to: show that a line meets a circle (substitute into the equation and solve a quadratic); find the midpoint of a chord using the perpendicular from the centre; find lengths with Pythagoras; and find areas of triangles formed by the centre, chord ends or tangent intercepts. Always state the property you use in words (for example 'the angle in a semicircle is 90∘90^\circ') because a mark is often given for the reason. Give exact values (surds) unless a decimal is requested.

Exam tip

Sketch the circle, label the centre and the points, and mark the right angles before doing any algebra.

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Exam questions on Circle properties

  1. A circle has diameter ABAB, where A=(1,2)A=(1,2) and B=(9,8)B=(9,8). The point P=(8,1)P=(8,1) lies on the circle.
    Show that angle APBAPB is a right angle, and name the circle property you are using.2 marks
  2. A circle has centre C=(2,3)C=(2,3) and radius 55. The points P=(5,7)P=(5,7) and Q=(6,0)Q=(6,0) lie on the circle.
    Find the exact distance from CC to the chord PQPQ.2 marks
  3. The circle SS has equation (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25. The point T=(4,2)T=(4,2) lies on SS.
    Find an equation of the tangent to SS at TT, giving your answer in the form ax+by+c=0ax+by+c=0 where aa, bb and cc are integers.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).