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Circumcircle of a Right-angled Triangle

Maths: Edexcel A Level Pure Maths, topic 3.2.9

Circumcircle of a Right-Angled Triangle

The circumcircle of a right-angled triangle is easily found by looking at the midpoint of its hypotenuse.

Triangle in a semi-circle

Triangle in a semi-circle

  • There is a circle theorem that states that the angle of a triangle at the circumference of a semi-circle is always 90°.
  • This means that the hypotenuse of a right-angled triangle is the diameter of its circumcircle.
  • We can use this to find the equation of the circumcircle of the triangle.
How do you find the hypotenuse of the right-angled triangle?

How do you find the hypotenuse of the right-angled triangle?

  • To identify which two vertices of the triangle form the hypotenuse, we first need to find the square of the length of each side.
  • We can identify the hypotenuse using Pythagoras' theorem:
  • Opposite side2 + adjacent side2 = hypotenuse2
How do you find the equation of the circumcircle?

How do you find the equation of the circumcircle?

  • To write the general equation of the circumcircle, we need to find the centre and the radius of the circle.
  • The centre is the midpoint of the hypotenuse.
  • The radius is half the length of the hypotenuse.
Example

Example

  • The points A(3,6), B(3,4) and C(4,5) lie on the circumference of a circle.
  • Show that ABC is a right-angled triangle.
  • Find the equation of the circle.
Work out the length of each side squared

Work out the length of each side squared

  • AB2=22=4AB^2 = 2^2 = 4
  • AC2=12+12=2AC^2 = 1^2 + 1^2 = 2
  • BC2=12+12=2BC^2 = 1^2 + 1^2 = 2
Use Pythagoras' theorem

Use Pythagoras' theorem

  • All right-angled triangles satisfy Pythagoras' theorem.
  • We can see that AC2+BC2=AB2AC^2 + BC^2=AB^2 as 2 + 2 = 4.
  • This means that ABCABC is a right-angled triangle with ABAB as its hypotenuse.
Use circle theorem

Use circle theorem

  • ABCABC is a right-angled triangle, so ABAB is the diameter of its circumcircle.
Find the centre

Find the centre

  • The centre of the circumcircle is the midpoint of the diameter ABAB.
  • Midpoint = (3+32,6+42)=(3,5)(\frac{3+3}{2},\frac{6+4}{2}) = (3,5)
Find the radius

Find the radius

  • As calculated earlier, AB2AB^2 = 4. This means:
  • Diameter ABAB = 2
  • Radius = 1
Write the equation of the circle

Write the equation of the circle

  • We can now write the equation of the circumcircle in its general form:
  • (x−3)2+(y−5)2=1(x-3)^2 + (y-5)^2 = 1

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