We already saw that $x=\cos(t)$, $y=\sin(t)$ gives a circle traced counter-clockwise. On this page, we'll see how to modify this curve to give circles and ellipses centered at arbitrary points.
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Example 1: Find a parametrization for a circle of radius 17 centered
at the origin, traced counterclockwise starting at the right.
Solution: Just use the parametrization of the unit circle (traced counterclockwise starting at the right) and multiply both $x$ and $y$ by 17 to make things bigger: $$ x(t) = 17 \cos(t); \qquad y(t) = 17 \sin(t).$$ |
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Example 2: Now find a parametrization
for a circle of radius 17, centered
at the origin, traced clockwise from the right.
Solution: Take the previous example and turn it upside down. This means replacing $y$ with $-y$, so now $$ x(t) = 17 \cos(t); \qquad y(t) = -17 \sin(t).$$ |
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Example 3: Find a parametrization for an ellipse, three times wider
and twice as tall as the unit circle, centered at the origin.
Solution: To make something three times wider, just multiply $x$ by 3, and to make something twice as tall, multiply $y$ by 2: $$ x(t) = 3\cos(t); \qquad y(t) = 2 \sin(t).$$ Note that this parametrization goes counterclockwise, starting at the right. How would you adjust the parametrization to go clockwise, starting at the left? |
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Example 4: Find a parametrization for a circle of radius 5 centered
at $(12,7)$.
Solution: Start with your favorite parametrization of the unit circle. Multiply $x$ and $y$ by 5 to get a circle of radius 5, still centered at the origin. To move 12 steps to the right and 7 steps up, just add 12 to $x$ and 7 to $y$. Put together, this yields $$ x(t) = 5\cos(t) + 12; \qquad y(t) = 5\sin(t) + 7.$$ |
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Example 5: Find a parametrization for an ellipse,
3 times as wide and 5 times as high as the unit circle, centered at
(-3,8).
Solution: You can work this one out yourself! |
| General case: The parametrized curve $$ x(t) = a \cos(t) + h; \qquad y(t) = b \sin(t) + k, $$ where $a$, $b$, $k$, and $h$ are constants, gives an ellipse of width $|a|$, height $|b|$, and center at $(h,k)$. If $a$ and $b$ are positive, then this is traced counterclockwise starting at the right. If $a<0$, then we start at the left, and if $ab<0$ then we go clockwise instead of counterclockwise. You should check for yourself that all of the previous examples fit into this pattern. |