Math Detours · Grade 9 · Around the corner · 18

The circle through three points

Any three points not in a line lie on exactly one circle. Its centre is where two perpendicular bisectors cross, and the third always passes through the same spot.

Name ____________________Date ____________

Tonight's idea

ABCO
Two bisectors (blue) fix \(O\). The third (orange) has to pass through it.

Take two points \(A\) and \(B\). The points equally far from both make a line. It passes through the midpoint of \(AB\) at right angles to \(AB\): the perpendicular bisector from Ruler and compass. Fold the page along it and \(A\) lands on \(B\), so every point on the fold is equally far from both.

Now take three points \(A\), \(B\), \(C\), not in a line. A circle through all three needs a centre \(O\) with \(OA = OB = OC\).

\(OA = OB\) puts \(O\) on the bisector of \(AB\). \(OB = OC\) puts \(O\) on the bisector of \(BC\). \(AB\) and \(BC\) point in different directions, so these two lines are not parallel and cross at exactly one point. That point is \(O\).

The third bisector? At \(O\), \(OA = OB\) and \(OB = OC\), so \(OA = OC\). That is the test for lying on the bisector of \(AC\). The third line has no choice.

Three points not in a line lie on exactly one circle. Its centre is where the perpendicular bisectors meet.

Watch one: the circle through \(A(0, 0)\), \(B(8, 0)\), \(C(7, 7)\)

\(AB\) is horizontal with midpoint \((4, 0)\), so its bisector is the vertical line \(x = 4\). The centre is \(O(4, k)\) for some \(k\).

Now use \(OA = OC\). Squared distances (Pythagoras) avoid square roots:

\[4^2 + k^2 = 3^2 + (7 - k)^2 \quad\Rightarrow\quad 16 + k^2 = 58 - 14k + k^2 \quad\Rightarrow\quad k = 3.\]

The centre is \((4, 3)\) and the radius is \(OA = \sqrt{16 + 9} = 5\). Check \(B\): \(4^2 + 3^2 = 25\). Check \(C\): \(3^2 + 4^2 = 25\). This is the circle in the figure.

Your turn

  1. \(A(0, 0)\), \(B(6, 0)\), \(C(0, 8)\). Two sides lie along the axes, so you can write two bisectors at sight. Find the centre and radius. Where does the centre sit on the triangle?
  2. Find the centre and radius of the circle through \(A(-2, -2)\), \(B(4, -2)\) and \(C(5, 5)\). Check your centre against all three points.
  3. \(A(0, -4)\), \(B(0, 10)\), \(C(6, 4)\). (a) Find the centre and the exact radius. (b) Is \(D(4, 8)\) on the circle? Is \(E(6, 3)\)?
  4. (a) There is no circle through \((0, 0)\), \((2, 2)\) and \((5, 5)\). What do the perpendicular bisectors do this time? (b) Two different circles are drawn on a page. What is the largest number of points they can share? Use tonight's idea to explain why it cannot be more.

Stretch: the arch

A stone arch is part of a circle. It is 8 m wide at ground level and 2 m high in the middle. Find the radius of the circle, and say where its centre is. (Put the feet of the arch at \((-4, 0)\) and \((4, 0)\).)

Answers

  1. The bisector of \(AB\) is \(x = 3\) and the bisector of \(AC\) is \(y = 4\). Centre \((3, 4)\), radius \(\sqrt{9 + 16} = 5\). The centre is the midpoint of \(BC\), the side opposite the right angle. (Next year, in Circle theorems, with proofs, you will see why that always happens.)
  2. \(AB\) is horizontal with midpoint \((1, -2)\), so the centre is \((1, k)\). \(OA = OC\): \(9 + (k + 2)^2 = 16 + (k - 5)^2\), so \(13 + 4k = 41 - 10k\) and \(k = 2\). Centre \((1, 2)\). Radius: \(OA^2 = 9 + 16 = 25\), so 5. Check \(B\): \(9 + 16 = 25\). Check \(C\): \(16 + 9 = 25\).
  3. (a) \(AB\) is vertical with midpoint \((0, 3)\), so the centre is \((h, 3)\). \(OA = OC\): \(h^2 + 49 = (h - 6)^2 + 1\), so \(49 = 37 - 12h\) and \(h = -1\). Centre \((-1, 3)\), radius \(\sqrt{1 + 49} = \sqrt{50} = 5\sqrt2\). (b) \(D\): \(5^2 + 5^2 = 50\). On the circle. \(E\): \(7^2 + 0^2 = 49\), less than 50. Not on it: \(E\) is just inside.
  4. (a) The three points lie on the line \(y = x\). Every bisector is perpendicular to that same line, so the bisectors are parallel and never meet. No point is equally far from all three. (b) Two. If two circles shared three points, those points could not be in a line (a line meets a circle at most twice), so exactly one circle passes through them. The two circles would be the same circle.

Stretch. The top of the arch is \((0, 2)\). The bisector of the two feet is the vertical line \(x = 0\), so the centre is \((0, k)\). Equal distances to a foot and to the top: \(16 + k^2 = (2 - k)^2 = 4 - 4k + k^2\), so \(k = -3\). The centre is 3 m below the ground, under the middle of the arch, and the radius is \(2 - (-3) = 5\) m. Check a foot: \(\sqrt{16 + 9} = 5\).

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