Question

Find the center of the circle defined by the equation.
x^2+y^2+16x+10y+53=0

standard form of the equation of a circle is:
(x-h)^2+(y-k)^2=r^2

Answer 1

Complete the square.

Answer 2

complete the square to write it in that form

Answer 3

do you know how to complete the square?

Answer 4

No

Answer 5

\[x^2+16x+y^2+10y=-53\]

Answer 6

i just grouped the terms together. now half of 16 is 8 and half of 10 is 5 so write
\[(x+8)^2+(y+5)^2=-53+8^2+5^2\]

Answer 7

the reason this works is that
\[(x+8)^2=x^2+8x+16\] so you have added 16 to the left and therefore have to add 16 to the right

Answer 8

check this:
(x+8)^2 -64+ (y+5)^2 -25 +53=0
(x+8)^2 +(y+5)^2=36
here is your center:
(-8,-5)

Answer 9

likewise
\[(x+5)^2=x^2+10x+25\] so you are adding 25 when you rewrite in that fashion. adjust by adding 25 to the right

Answer 10

Is it (3,5)

Answer 11

no center is (-3,-5)

Answer 12

not that either

Answer 13

(-8,-5) ?

Answer 14

you get
\[(x+8)^2+(y+5)^2=36\] yes?

Answer 15

so center is (-8,-5) and radius is 6

Answer 16

that's that i put up there... :)

Answer 17

x^2 + y^2 - 6x - 10y + 30 = 0
..
x^2-6x +y^2-10y+30=0
(x^2-6x+9)-9+(y^2-10y+25)-25+30=0
(x-3)^2+(y-5)^2-9-25+30=0
(x-3)^2+(y-5)^2=9+25-30
(x-3)^2+(y-5)^2=4
Center = (3,5)
sum = 8

Answer 18

this may be correct but it is not the problem up top!

Answer 19

agreed with satellite73 :)

Answer 20

I I messed the 16 up! uggghh.

Answer 21

btw i find it easier to just put all the numbers on the right hand side instead of adding and subtracting. of course it doesn't matter, it is just my personal choice.

Answer 22

in other words if i see
\[x^2+10x-3=0\] i would rewrite as
\[x^2+10x=3\] and then put
\[(x+5)^2=3+25