# centre semi circle

• Feb 21st 2010, 05:46 AM
wolfhound
centre semi circle
Hello
if y= squareroot R^2 -x^2
how does that justify the centre coordinates is (0,0) ?
• Feb 21st 2010, 06:18 AM
Quote:

Originally Posted by wolfhound
Hello
if y= squareroot R^2 -x^2
how does that justify the centre coordinates is (0,0) ?

Hi wolfhound,

Remember the "distance formula" ?

$\displaystyle distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^3}$

This is Pythagoras' Theorem using co-ordinates.

Just as on a clock, the hands are pivoted at the centre,
the hands are always the same length, but the point moves around the clock circumference. The circle circumference is the set of all points that are the same distance from the centre.

If the centre co-ordinates are (0,0), this reduces to

$\displaystyle r=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r^2=x^2+y^2$

$\displaystyle y^2=r^2-x^2$

$\displaystyle y=\sqrt{r^2-x^2}$
• Feb 21st 2010, 06:24 AM
wolfhound
Hi Archie
Great explanation, Thanks!