how do i express $\displaystyle x^2+y^2-28x+4y+120=0$ in parametric form
Hello, sigma1!
How do i express $\displaystyle x^2+y^2-28x+4y+120\:=\:0$ in parametric form?
We have: .$\displaystyle x^2 - 28x + y^2 + 4y \;=\;-120$
Complete the square: .$\displaystyle x^2 - 28x {\color{blue}\:+\: 196} + y^2 + 4y {\color{red}\:+\: 4} \;=\;-120 {\color{blue}\:+\: 196} {\color{red}\:+\: 4}$
And we have: .$\displaystyle (x-14)^2 + (y+2)^2 \;=\;80$
This is a circle with center $\displaystyle (14,-2)$ and radius $\displaystyle 4\sqrt{5}$
The parametric equations are: .$\displaystyle \begin{Bmatrix}x &=& 14 + 4\sqrt{5}\,\cos\theta \\ \\[-3mm] y &=& \text{-}2 + 4\sqrt{5}\,\sin\theta \end{Bmatrix}$
In general,
$\displaystyle x^2+y^2 +ax+by +c=0$
$\displaystyle x^2+ 2\frac a2x + (\frac a2)^2+y^2+2\frac b2y +(\frac b2)^2+c-(\frac a2)^2-(\frac b2)^2=0$
$\displaystyle (x-\frac a2)^2+(y-\frac b2)^2=(\frac a2)^2+(\frac b2)^2-c$
which you can check by expanding the squares. In your case, a=-28, b=4, c=120.
http://www.algebralab.org/lessons/le...gthesquare.xml
Oh. A circle of radius r centred at the origin is given by the parametric equations
$\displaystyle \begin{Bmatrix}x &=& r\,\cos\theta \\ \\[-3mm] y &=& r\,\sin\theta \end{Bmatrix}$.
This is pretty much just a definition. Can you translate these so that the centre is at the correct point?
a circle centered at the origin, $\displaystyle x^2 + y^2 = r^2$ , can be written parametrically as
$\displaystyle x = r\cos{t}$
$\displaystyle y = r\sin{t}$
shifting the center from the origin to the point $\displaystyle (h,k)$ yields the equations
$\displaystyle x = h + r\cos{t}$
$\displaystyle y = k + r\sin{t}$
... that's all.