Question

Need some help...

Answer 1

please, where is your question?

Answer 2

Where are you stuck?

Answer 3

How would I solve this system of equations? I know that they are both circles equal to 0.

Answer 4

I would first compelte the square for each function to put them in the form: \((x-h)^2 +(y-k)^2 =r^2\)

Answer 5

ok so I would get (x+1)^2 + (y+1)^2 and (x+2)^2 = (y+3)^2 -1

Answer 6

equating your equations each other, we find that:
\[x ^{2}+y ^{2}+2x+2y=x ^{2}+y ^{2}+4x+6y+12\]
and some simplification you should get this:
\[x+2y+6=0\]
that is the equation of the radical axis of your circumferences, namely:

Answer 7

first equation:\[x^2+y^2 +2x+2y=0\]\[\implies(x^2+2x)+(y^2+2y)=0\]\[\implies (x^2+2x+\color{red}1) +(y^2+2y+\color{red}1) -1-1=0\]\[\implies (x+1)^2+(y+1)^2-2=0\]\[\implies (x+1)^2+(y+1)^2 =2\]

Answer 8

since radical axis passes at intersection points, you have to solve this system, please:
\[x ^{2}+y ^{2}+2x+2y=0,x+2y+6=0\]

Answer 9

Second equation: \[x^2+y^2 +4x+6y+12=0\]\[\implies (x^2+4x)+(y^2+6y)+12=0\]\[\implies (x^2+4x+\color{red}{4} )+(y^2 +6y +\color{red}{9})+12-4-9=0\]\[\implies (x+2)^2 +(y+3)^2 -1=0\]\[\implies (x+2)^2 +(y+3)^2 =1\]I think @Michele_Laino's method is a bit simpler and easier to follow though.

Answer 10

thanks! @Jhannybean