Hello

if y= squareroot R^2 -x^2

how does that justify the centre coordinates is (0,0) ?

someone explain please

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- Feb 21st 2010, 05:46 AMwolfhoundcentre semi circle
Hello

if y= squareroot R^2 -x^2

how does that justify the centre coordinates is (0,0) ?

someone explain please - Feb 21st 2010, 06:18 AMArchie Meade
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 AMwolfhound
Hi Archie

Great explanation, Thanks!