Circles, parametric curves and broken graphs

The problem

Ask a shopkeeper the price of 1 kg of sugar and you get one answer. Ask a class who is 16 years old and several hands go up. The first behaves like a function: one input, one output. The second does not.

A circle is like the second question: one x can give two y-values. This lesson graphs such curves, curves given through a parameter t, and graphs that break.

x² + y² = 4 is not a function of x, because a vertical line can cut it twice. It is the union of two functions, y = √(4 − x²) and y = −√(4 − x²). To graph any such curve, check symmetry, find intercepts, make a table and join the points smoothly.

Key terms

Implicit equation

x and y appear together, not as y = f(x). Solving for y may give ±, meaning two y-values for one x.

Symmetry test

If replacing x by −x leaves the equation unchanged, the graph is symmetric about the y-axis. Replacing y by −y tests the x-axis.

Parametric equations

x and y are both given in terms of a third variable t. Each t gives one point (x, y); eliminating t gives the curve's equation.

Discontinuous graph

A graph with a jump between pieces, or a single missing point. Mark an excluded point with an open circle.

Short questions with model answers

1. Q1. Graph the circle x² + y² = 4.

   • (x, y) → (−x, y), (x, −y), (−x, −y): no change
   • x = 0: y = ±2; x = 1: y = ±√3 = ±1.732; x = 2: y = 0

   A circle of radius 2 centred at the origin

2. Q2. Graph x = t², y = t for −2 ≤ t ≤ 2.

   • t = −2, −1, 0, 1, 2 → (4, −2), (1, −1), (0, 0), (1, 1), (4, 2)
   • t = y ⇒ x = y², i.e. y² = x

   y² = x

3. Q3. Graph y = (x² − 9)/(x − 3), x ≠ 3.

   • (x − 3)(x + 3)/(x − 3) = x + 3, x ≠ 3
   • Open circle at (3, 6)

   The line y = x + 3 with a hole at (3, 6)

Common mistakes

• ✗ Writing y = √(4 − x²) and drawing only the top half of the circle.

  ✓ y² = 4 − x² gives y = ±√(4 − x²). The minus sign gives the bottom half; you need both.

• ✗ Plotting (t, x) or (t, y) for parametric equations.

  ✓ t only labels the points. Plot (x, y): for t = −2, the point is (4, −2).

• ✗ Drawing y = (x² − 9)/(x − 3) as an unbroken line.

  ✓ At x = 3 the formula gives 0/0, so there is no point. Draw an open circle at (3, 6).

MCQs

1. 1. On x² + y² = 4, the points with x = 1 have y equal to:

   1. (a) ±1
   2. (b) ±√3
   3. (c) ±2
   4. (d) √3 only
   Show answer

   (b) y² = 4 − 1 = 3, so y = ±√3 ≈ ±1.732: two points.

2. 2. The y-intercepts of 9x² + 4y² = 36 are:

   1. (a) ±2
   2. (b) ±3
   3. (c) ±4
   4. (d) ±9
   Show answer

   (b) Put x = 0: 4y² = 36, so y² = 9 and y = ±3.

3. 3. Eliminating t from x = t², y = t gives:

   1. (a) y = x²
   2. (b) y² = x
   3. (c) x² + y² = 1
   4. (d) y = 2x
   Show answer

   (b) y = t, so x = t² = y².

4. 4. For y = x (0 ≤ x ≤ 1) and y = x − 1 (1 < x ≤ 2), the value of y at x = 1.5 is:

   1. (a) 1.5
   2. (b) 0.5
   3. (c) 1
   4. (d) 2.5
   Show answer

   (b) 1.5 lies in 1 < x ≤ 2, so use y = x − 1 = 0.5.

5. 5. The graph of y = (x² − 4)/(x − 2), x ≠ 2, has a hole at:

   1. (a) (2, 0)
   2. (b) (2, 4)
   3. (c) (−2, 0)
   4. (d) (0, 2)
   Show answer

   (b) It is y = x + 2 with x ≠ 2, so the missing point is (2, 2 + 2) = (2, 4).

Quick revision

• An implicit curve like x² + y² = 4 fails the vertical line test, but splits into two functions with + and − roots.
• For parametric equations, tabulate t, plot (x, y), and eliminate t to name the curve.
• Discontinuous graphs show jumps between pieces or holes at excluded points; mark those with open circles.