Circle & Ellipse

Circle

$$ \boxed{ \text{General formula: } {(x - h)^2 \over r^2} + {(y - k)^2 \over r^2} = 1 } $$

Features:

• Centre $(h, k)$ and radius of $r$ units
• $x$-values restricted to $h - r \le x \le h + r$
• $y$-values restricted to $k - r \le y \le k + r$

$$ {(x - 2)^2 \over 3^2} + {(y + 1)^2 \over 3^2} = 1 $$

Features:

• Centre $(2, -1)$ and radius of $3$ units
• $x$-values restricted to $-1 \le x \le 5$
• $y$-values restricted to $-4 \le y \le 2$

Ellipse

$$ \boxed{ \text{General formula: } {(x - h)^2 \over a^2} + {(y - k)^2 \over b^2} = 1 } $$

Features:

• Centre $(h, k)$, horizontal width $2a$ and vertical width $2b$
• $x$-values restricted to $h - a \le x \le h + a$
• $y$-values restricted to $k - b \le y \le k + b$

$$ {(x - 2)^2 \over 3^2} + {y^2 \over 2^2} = 1 $$

Features:

• Centre $(2, 0)$, horizontal width $6$ units and vertical width $4$ units
• $x$-values restricted to $-1 \le x \le 5$
• $y$-values restricted to $-2 \le y \le 2$

Change equation to general form

$$ 4x^2 + y^2 + 16x + 7 = 0 $$

To change from the above to the ${(x -h)^2 \over a^2} + { (y - k)^2 \over b^2} = 1$ form, complete the square for $x$ and manipulate:

\begin{align} 4x^2 + y^2 + 16x + 7 & = 0 \\ 4x^2 + 16x + y^2 & = - 7 \\ x^2 + 4x + {y^2 \over 4} & = -{7 \over 4} \\ [\text{Complete the square}] \phantom{00000} (x + 2)^2 - (2)^2 + {y^2 \over 4} & = -{7 \over 4} \\ (x + 2)^2 - 4 + {y^2 \over 4} & = -{7 \over 4} \\ (x + 2)^2 + {y^2 \over 4} & = -{7 \over 4} + 4 \\ (x + 2)^2 + {y^2 \over 4} & = {9 \over 4} \\ 4(x + 2)^2 + y^2 & = 9 \\ {4(x + 2)^2 \over 9} + {y^2 \over 9} & = 1 \\ { (x + 2)^2 \over {9 \over 4} } + {y^2 \over 9} & = 1 \\ { (x + 2)^2 \over \left(3 \over 2\right)^2 } +{y^2 \over 3^2} & = 1 \end{align}

The equation represents an ellipse with centre $(-2, 0)$, horizontal width $3$ units and vertical width $6$ units.

'Top' & 'Bottom' Half, 'Left' & 'Right' Half

$$ {(x - 2)^2 \over 3^2} + {y^2 \over 2^2} = 1 $$

The equation above represents an ellipse with centre $(2, 0)$, horizontal width 6 units and vertical width 4 units.

To obtain the equation for the 'top' half and 'bottom' half, make $y$ the subject:

\begin{align} {(x - 2)^2 \over 3^2} + {y^2 \over 2^2} & = 1 \\ {(x - 2)^2 \over 9} + {y^2 \over 4} & = 1 \\ {y^2 \over 4} & = 1 - {(x - 2)^2 \over 9} \\ {y^2 \over 4} & = 1 - {1 \over 9} (x - 2)^2 \\ y^2 & = 4 - {4 \over 9} (x - 2)^2 \\ y & = \pm \sqrt{4 - {4 \over 9}(x - 2)^2 } \end{align}

$$ {(x - 2)^2 \over 3^2} + {y^2 \over 2^2} = 1 $$

On the other hand, to obtain the equation for the 'left' half and 'right' half, make $x$ the subject:

\begin{align} {(x - 2)^2 \over 3^2} + {y^2 \over 2^2} & = 1 \\ {(x - 2)^2 \over 9} + {y^2 \over 4} & = 1 \\ {(x - 2)^2 \over 9} & = 1 - {y^2 \over 4} \\ (x - 2)^2 & = 9 - {9 \over 4}y^2 \\ x - 2 & = \pm \sqrt{9 - {9 \over 4}y^2} \\ x & = \pm \sqrt{9 - {9 \over 4}y^2} + 2 \end{align}