## anonymous 4 years ago Find the equation of a circle with centre (0,0) that passes through (8,-15)

1. anonymous

x^2 + y ^2 = r ^2 (substitute the values :D

2. anonymous

can yu show me plug in the values

3. anonymous

$x^2 + y ^2 = r ^2$ the length of the radius of the circle is $r=\sqrt{(8)^2+(-15)^2} = 17$ $x^2+y^2=289$ <-- is the equation of the circle

4. anonymous

lol fine -.- (8)^2 + (-15)^2 = r^2 64 + 225 = r ^2 289 = r^2 13 = r

5. anonymous

where did yu get 17 from

6. anonymous

its 13 not 17

7. anonymous

nvm its 17

8. anonymous

how do yu kno

9. anonymous

lol kay hes right its 17

10. anonymous

he took the square root of 17

11. anonymous

i mean 289

12. anonymous

yeh, just look at the answer up there and you will know where the 17 came from, it is the magnitued of the radius, whenever you want to find the magnited of a vector lets say (2,3), it is equal to the square root of the x, y values square. take a look at this: http://www.mathsisfun.com/algebra/circle-equations.html

13. anonymous

), it is equal to the square root of the x, y values SQUARED**