Whole-number points on a circle
Tonight's idea
Draw the circle with centre \((0, 0)\) and radius 5. By Pythagoras, a point \((x, y)\) is at distance 5 from the centre exactly when \(x^2 + y^2 = 25\). So the circle and the equation are one object seen two ways: a point is on the circle if its coordinates fit the equation, and not otherwise.
That turns questions about the shape into algebra. Is \((3, 4)\) on the circle? Check: \(9 + 16 = 25\), so yes. Where does a line cross the circle? A crossing point must fit both equations at once, so substitute the line into the circle and solve.
Substituting a straight line into a circle leaves a quadratic in one letter. Two solutions: the line crosses twice. One: it touches. None: it misses.
Studying shapes through their polynomial equations is called algebraic geometry. It is also how a computer handles shapes. A program does not look at the picture. It solves the equations.
Watch one: every whole-number point on \(x^2 + y^2 = 25\)
Here "whole-number" means any integer, so negatives and 0 count. If \(x\) is 6 or more, \(x^2\) alone is over 25, so try \(x = 0\) to 5 and ask whether \(25 - x^2\) is a square.
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \(25 - x^2\) | 25 | 24 | 21 | 16 | 9 | 0 |
| \(y\) | \(\pm 5\) | none | none | \(\pm 4\) | \(\pm 3\) | 0 |
Now let \(x\) be negative too. That gives \((0, \pm 5)\) and \((\pm 5, 0)\), 4 points, then \((\pm 3, \pm 4)\) and \((\pm 4, \pm 3)\) with every choice of signs, 8 points. Total: 12.
A line. Where does \(y = x + 1\) meet the circle? Substitute: \(x^2 + (x + 1)^2 = 25\), so \(2x^2 + 2x - 24 = 0\), so \(x^2 + x - 12 = 0\), so \((x + 4)(x - 3) = 0\). Then \(x = 3\) gives \(y = 4\), and \(x = -4\) gives \(y = -3\). The line crosses at \((3, 4)\) and \((-4, -3)\), and nowhere else.
Your turn
- Find every whole-number point on the circle \(x^2 + y^2 = 50\). Make a table for \(x = 0\) to 7, then put in the signs. How many points are there?
- Show that the circle \(x^2 + y^2 = 3\) passes through no whole-number points at all.
- Where does the line \(x + y = 7\) meet the circle \(x^2 + y^2 = 25\)? Write \(y = 7 - x\), substitute and solve.
- Both parts use the circle \(x^2 + y^2 = 25\). (a) Show that the line \(y = x + 10\) misses the circle. (b) The line \(3x + 4y = 25\) can be written \(y = \tfrac{25 - 3x}{4}\). Substitute, multiply through by 16, and solve. How many points do you get, and what does that say about the line?
Stretch: points with fraction coordinates
The point \((-5, 0)\) is on the circle \(x^2 + y^2 = 25\). A line through it with slope \(m\) is \(y = m(x + 5)\). (a) Take \(m = \tfrac12\). Where else does the line hit the circle? (b) Take \(m = \tfrac15\). The second point now has fraction coordinates. Find it and check that it fits the equation. (c) Why must every fraction slope give a second point whose coordinates are fractions (or whole numbers)?
Answers
- For \(x = 0, 1, \dots, 7\) the values of \(50 - x^2\) are 50, 49, 46, 41, 34, 25, 14, 1. Only 49, 25 and 1 are squares, at \(x = 1, 5, 7\). With signs: \((\pm 1, \pm 7)\), \((\pm 5, \pm 5)\) and \((\pm 7, \pm 1)\), 4 points each, so 12 points. If \(x\) is 8 or more, \(x^2\) is already over 50.
- If \(x\) is 2 or more, or \(-2\) or less, then \(x^2\) is at least 4, which is over 3. So \(x\) is \(-1\), 0 or 1 and \(x^2\) is 0 or 1. Then \(y^2\) would be 3 or 2, and neither is the square of a whole number. The circle is real, with radius \(\sqrt3 \approx 1.73\), and it has plenty of points. It just slips between all the grid points.
- \(x^2 + (7 - x)^2 = 25\) gives \(2x^2 - 14x + 49 = 25\), so \(2x^2 - 14x + 24 = 0\), so \(x^2 - 7x + 12 = 0\), so \((x - 3)(x - 4) = 0\). Then \(x = 3, y = 4\) or \(x = 4, y = 3\). The line crosses at \((3, 4)\) and \((4, 3)\).
- (a) \(x^2 + (x + 10)^2 = 25\) gives \(2x^2 + 20x + 75 = 0\). Complete the square: the left side is \(2(x + 5)^2 + 25\). A square is never negative, so this is at least 25 and can never be 0. No solutions, so no point is on both. (b) \(x^2 + \tfrac{(25 - 3x)^2}{16} = 25\) becomes \(16x^2 + 625 - 150x + 9x^2 = 400\), so \(25x^2 - 150x + 225 = 0\), so \(x^2 - 6x + 9 = 0\), so \((x - 3)^2 = 0\). The only solution is \(x = 3\), with \(y = \tfrac{25 - 9}{4} = 4\). One point, \((3, 4)\): the line touches the circle there. It is a tangent.
Stretch. (a) \(y = \tfrac{x + 5}{2}\), so \(x^2 + \tfrac{(x + 5)^2}{4} = 25\). Multiply by 4: \(5x^2 + 10x - 75 = 0\), so \(x^2 + 2x - 15 = 0\), so \((x + 5)(x - 3) = 0\). The new point is \((3, 4)\). (b) \(y = \tfrac{x + 5}{5}\), so \(25x^2 + (x + 5)^2 = 625\), so \(13x^2 + 5x - 300 = 0\), so \((x + 5)(13x - 60) = 0\). The new point is \(\left(\tfrac{60}{13}, \tfrac{25}{13}\right)\). Check: \(60^2 + 25^2 = 4225 = 25 \times 13^2\). (c) The quadratic has fraction coefficients, and \(x = -5\) is always one solution, so \((x + 5)\) is a factor. The other factor is a straight-line expression with fraction coefficients, so its solution is a fraction, and then \(y = m(x + 5)\) is a fraction too. In general the second point is \(\left(\tfrac{5(1 - m^2)}{1 + m^2}, \tfrac{10m}{1 + m^2}\right)\). Different slopes give different points, so this circle has only 12 whole-number points but infinitely many fraction points.