Question

I am supposed to identify the center and radius of each circle given the equation. My equations are:
x^2+y^2+8x-8y+31=0 and
x^2+y^2+8x+7=0

Answer 1

you have to plug that into the general formula for the circle of center (h,k) \[(x-h)^2+(y-k)^2=r^2\]

Answer 2

you need to factorise something from it, we know that \[(a-b)^2=(a-b)(a-b)=a^2-2ab-b\]

Answer 3

sorry b^2

Answer 4

so we can force the equation to give us such values like this:
we have the \[x^2+y^2+8x-8y+31\] and we take the \[x^2+8x\] to factorise as part of the \[(x-h)^2\] so we can see that \[x^2-2hx+h^2=x^2+8x+16\] and we make that into \[(x+4)^2\] and we have to take that 16 units we used for our x from the 31 so we get this \[(x+4)^2+y^2-8y+15=0\]

Answer 5

then we have to get our \[(y-k)^2\] the same way

Answer 6

so we part from \[(x+4)^2+y^2-8y+15=0\] to get \[(y-k)^2=y^2-2ky+k^2=y^2-8y+16=(y-4)^2\]
and doing the same thing we did with the x, we need to take those 16 units from the rest and we get something like this \[(x+4)^2+(y-4)^2-1=0\]

Answer 7

and from there we get the general equation of \[(x-h)^2+(y-k)^2=r^2;(x+4)^2+(y-4)^2=1\]
so the centre is\[(h,k)=(-4,4)\] and the radius is\[\sqrt{1}=1\]

Answer 8

in the other one we have something really similiar, but this time is \[(x+4)^2+y^2=9\] so\[(h,k)=(-4,0)\] and \[r=\sqrt{9}=3\]