Complete the square to rewrite the following equation. Identify the center and radius of the circle. You must show all work and calculations to receive credit.

x2 + 4x + y2 − 6y = −4

Question

Complete the square to rewrite the following equation. Identify the center and radius of the circle. You must show all work and calculations to receive credit.

x2 + 4x + y2 − 6y = −4

crispyrat · Answer

is there a screenshot? there is not enough info ;-;

loganisbored · Answer

yes there is

crispyrat · Answer

no not really
for example x2 i don't know what that means, that could be a mistype of x*2 or x^2

crispyrat · Answer

so im just gonna assum its x^2
x^2+4x+y^2-6y=4
the equation should be
(a+b)(a+b)=a^2+2*a*b+b^2
so in this case we have to do that for both
x^2+4x so half of 4 is 2 so the thing is (x+2)(x+2) if we get x^2+4x+4 so we add 4 to -4
then  y^2 − 6y=(y-3)(y-3)= y^2 − 6y+9
so then we add the 9 to -4 as well
so we have
(x+2)^2+(y-3)^2=-4+4+9
(x+2)^2+(y-3)^2=9

Then the center is (-x,-y) and the radius is sqrt((-x)^2+(-y)^2)
so using this what is the radius and the center ;-;

crispyrat · Answer

@AZ can u check my work ;-;

@crispyrat wrote:

so im just gonna assum its x^2 x^2+4x+y^2-6y=4 the equation should be (a+b)(a+b)=a^2+2*a*b+b^2 so in this case we have to do that for both x^2+4x so half of 4 is 2 so the thing is (x+2)(x+2) if we get x^2+4x+4 so we add 4 to -4 then y^2 − 6y=(y-3)(y-3)= y^2 − 6y+9 so then we add the 9 to -4 as well so we have (x+2)^2+(y-3)^2=-4+4+9 (x+2)^2+(y-3)^2=9 Then the center is (-x,-y) and the radius is sqrt((-x)^2+(-y)^2) so using this what is the radius and the center ;-;

loganisbored · Answer

your smart

crispyrat · Answer

stop spamming. idrc what you say but don't spam on a question with unneeded compliments

loganisbored · Answer

im not

loganisbored · Answer

i said one thing and it was a compliment so chill out and stop complaining about one little thing

snowflake0531 · Answer

@crispyrat wrote:

so im just gonna assum its x^2 x^2+4x+y^2-6y=4 the equation should be (a+b)(a+b)=a^2+2*a*b+b^2 so in this case we have to do that for both x^2+4x so half of 4 is 2 so the thing is (x+2)(x+2) if we get x^2+4x+4 so we add 4 to -4 then y^2 − 6y=(y-3)(y-3)= y^2 − 6y+9 so then we add the 9 to -4 as well so we have (x+2)^2+(y-3)^2=-4+4+9 (x+2)^2+(y-3)^2=9 Then the center is (-x,-y) and the radius is sqrt((-x)^2+(-y)^2) so using this what is the radius and the center ;-;

You're all correct just one thing, radius is the r in $$(x-h)^2+(y-k)^2=r^2$$ and the center is (h,k) I was a bit confused when you said "Then the center is (-x,-y) and the radius is sqrt((-x)^2+(-y)^2)"

snowflake0531 · Answer

Also, the standard form of the circle should be written in r^2 so
$$(x+2)^2+(y-3)^2=3^2$$

crispyrat · Answer

ah ok i must have misworded somethingsit was a long time since i used quadratics