find a formula for the following function
the top half of a circle with center (a,b) and radius r
need some help with this one thanks...
Every point $\displaystyle (x,y)$ on the graph of a circle of radius $\displaystyle r$ satisfies the following relationship:
$\displaystyle x^2+y^2 = r^2 $
So to express $\displaystyle y$ as a function of $\displaystyle x$ rewrite:
$\displaystyle y = \pm \sqrt{r^2-x^2}$
The top half of the circle is just $\displaystyle y = \sqrt{r^2-x^2}$
Now to shift this function to be centered at $\displaystyle (a,b)$ we have to shift the function up by $\displaystyle b$ and over by $\displaystyle a$.
$\displaystyle y = \sqrt{r^2-(x-a)^2}+b$
The equation of a circle with center (a,b) is given by
$\displaystyle (x-a)^2+(y-b)^2=r^2$
to find the top half we need to isolate y and take the positive square root
$\displaystyle (y-b)^2=r^2-(x-a)^2 $ taking the postive root
$\displaystyle y-b=\sqrt{r^2-(x-a)^2 }$
so we get...
$\displaystyle y=b+\sqrt{r^2-(x-a)^2 }$
The equation of a circle is:
$\displaystyle (x-h)^2 + (y-k)^2 = r^2$
Where (h,k) is the center of the circle.
The equation for the two halves of a circle are given by isolating y.
$\displaystyle y = \pm \sqrt {r^2 - (x-h)^2} + k$
We are only interested in the top half so we only take the positive square root:
$\displaystyle y = \sqrt {r^2 - (x-h)^2} + k$