Question

Find the standard form of the following equation:
x^2+y^2+4x+10y+28=0

Answer 1

Yes, I need to write the standard form and then graph the circle.

Answer 2

oops i meant
\[(x-h)^2+(y-k)^2=r^2\]

Answer 3

first rewrite as
\[x^2+4x+y^2+10y=-28\] then complete the square twice
do you know how to do that?

Answer 4

No i don't, I seem to be stuck and not understanding it

Answer 5

ok i can walk you through it, it is not too hard

Answer 6

just the mechanics. not the reasoning
lets look at \(x^2+4x\)
what is half of \(4\) ?

Answer 7

2

Answer 8

ok so we write \((x+2)^2\)
now what is \(2^2\) ?

Answer 9

4

Answer 10

ok good
so when we replace the \(x^2+4x\) on the left with \((x+2)^2\) we have to add \(4\) to the other side

Answer 11

\[(x+2)^2+y^2+10y=-28+4\] or
\[(x+2)^2+y^2+10y=-24\] now we repeat the process with \(y^2+10y\)

Answer 12

what is half of \(10\) ?

Answer 13

5

Answer 14

so we are going to replace \(y^2+10y\) by \((y+5)^2\)
now that is \(5^2\) ?

Answer 15

25

Answer 16

then when we replace \(y^2+10y\) by \((y+5)^2\) we have to add \(25\) to the other side just like before
now we have
\[(x+2)^2+(y+5)^2=-24+25\] or
\[(x+2)^2+(y+5)^2=1\] as we want

Answer 17

Wow you made it seem so simple, thank you! How do I graph it now?

Answer 18

it is a circle with center at \((-2,-5)\) and radius \(1\)
draw that

Answer 19

Thank you!

Answer 20

yw