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...

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- Mar 27th 2008, 10:08 AMmathleteCircle problem
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... - Mar 27th 2008, 10:23 AMiknowone
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 $\displaystyle a$.

$\displaystyle y = \sqrt{r^2-(x-a)^2}+b$ - Mar 27th 2008, 10:24 AMTheEmptySet
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...

$\displaystyle y=b+\sqrt{r^2-(x-a)^2 }$ - Mar 27th 2008, 10:32 AMtopher0805
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:

$\displaystyle y = \sqrt {r^2 - (x-h)^2} + k$