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Equation of a circleEdexcel International A Level Maths: Revision notes

Section 1

The equation of a circle

A circle is the set of points a fixed distance rr (the radius) from a fixed point (a,b)(a,b) (the centre). Using the distance formula, a point (x,y)(x,y) is on the circle when (x−a)2+(y−b)2=r\sqrt{(x-a)^2+(y-b)^2}=r, which gives (x−a)2+(y−b)2=r2.(x-a)^2+(y-b)^2=r^2. If the centre is the origin, this becomes x2+y2=r2x^2+y^2=r^2.

The right-hand side is r2r^2, so the radius is its square root.

Key termscirclecentreradius
Common mistake

Writing rr instead of r2r^2 on the right-hand side. For radius 44 the equation ends '=16=16'.

Section 2

Reading the centre and radius

From (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2, read off the centre (a,b)(a,b) and radius rr. The signs in the brackets are the opposite of the coordinates.

Example: (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25. Write (y+2)=(y−(−2))(y+2)=(y-(-2)), so the centre is (3,−2)(3,-2) and the radius is 25=5\sqrt{25}=5.

Example: (x+4)2+(y−1)2=13(x+4)^2+(y-1)^2=13. The centre is (−4,1)(-4,1) and the radius is 13\sqrt{13}.

Key termsstandard form
Common mistake

Reading the centre with the signs from the brackets: (y+2)(y+2) means yy-coordinate −2-2.

Section 3

Finding the equation from given information

To write the equation you need the centre (a,b)(a,b) and r2r^2.

Centre and a point on the circle. The radius is the distance from the centre to the point. Centre (−4,1)(-4,1) through (−2,4)(-2,4): r2=(−2+4)2+(4−1)2=4+9=13r^2=(-2+4)^2+(4-1)^2=4+9=13, so (x+4)2+(y−1)2=13(x+4)^2+(y-1)^2=13.

Diameter given by its end points. The centre is the midpoint of the diameter, and rr is half the length of the diameter. For A(−3,1)A(-3,1) and B(5,5)B(5,5) the centre is (1,3)(1,3) and r2=(5−1)2+(5−3)2=20r^2=(5-1)^2+(5-3)^2=20, so (x−1)2+(y−3)2=20(x-1)^2+(y-3)^2=20.

Key termsmidpointdiameter
Exam tip

Square the differences in xx and yy and add them. You get r2r^2, which is what the equation needs, so you rarely need a square root.

Section 4

Expanded form: completing the square

A circle may be given as x2+y2+px+qy+c=0x^2+y^2+px+qy+c=0. Complete the square in xx and in yy to find the centre and radius.

Example: x2+y2−10x+4y+13=0x^2+y^2-10x+4y+13=0. (x−5)2−25+(y+2)2−4+13=0(x-5)^2-25+(y+2)^2-4+13=0, so (x−5)2+(y+2)2=16(x-5)^2+(y+2)^2=16. The centre is (5,−2)(5,-2) and the radius is 44.

If the number on the right-hand side is not positive, there is no circle.

Key termscompleting the square
Common mistake

Forgetting to move the constants to the right-hand side: here 25+4−13=1625+4-13=16, not 25+4+1325+4+13.

Section 5

Points, lines and the circle

To find where a circle meets a line, substitute the line's equation into the circle's equation and solve.

For the circle above, the xx-axis is y=0y=0: (x−5)2+4=16(x-5)^2+4=16, so (x−5)2=12(x-5)^2=12 and x=5±23x=5\pm2\sqrt3.

To test a point (p,q)(p,q), calculate (p−a)2+(q−b)2(p-a)^2+(q-b)^2 and compare with r2r^2. If it is less than r2r^2 the point is inside the circle, if equal it is on the circle, and if greater it is outside.

Example: for (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25 the point (6,2)(6,2) gives 9+16=259+16=25, so it lies on the circle.

Key termsinsideoutside
Exam tip

The greatest and least values of xx and yy on a circle are a±ra\pm r and b±rb\pm r.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Equation of a circle

  1. A circle has equation (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25.
    Determine whether the point (6,2)(6,2) lies inside, outside or on the circle.2 marks
  2. A circle has centre (−4,1)(-4,1) and passes through the point (−2,4)(-2,4).
    Find the coordinates of the points where the circle meets the line y=1y=1.2 marks
  3. A circle CC has equation x2+y2−10x+4y+13=0x^2+y^2-10x+4y+13=0.
    Find the coordinates of the centre of CC and the radius of CC.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).