Algebra question?
Decide whether or not the equation has a circle as its graph. If it does, give the center a\[x^2+y^2-6x+2y=-9\]nd the radius. If it does not, describe the graph.
I know I need to rewrite into center radius form but i only get this far: \[(x^2-6x)+(y^2+2y)=-9\]
what is the general form for circle?
\[(x-h)^2+(y-k)^2=r^2\] ?
does yours look like that?
no:(
\[(x-6)^2+(y-2)^2=-9^2\]
can you expand the left hand side?
wait, it'd be x+6 right? not x-6
you complete the squares
\[(x+6)^2+(y-2)^2=-9^2\]
look at the link I shared it shows you step by step using similar problem
ok hang on
I am closing to finding the answer will you tell me if its right?
Ok, I got (3,-1) and now cant figure out to how find the radius?
@nincompoop
hmm https://www.youtube.com/watch?v=qh5hNY83UA4 is an accurate video that follows up the process
am I completely wrong? :/ I have trouble with longer problems. can anyone solve this exact problem so I can see where I went wrong?
Can anyone just give me answer at least? Trying to finish my homework up and this is my last problem left.
@nincompoop @dehelloo
AFK.
he died ?? ;-; nin!!
Haha I don't know what happened!
\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia \[(x+6)^2+(y-2)^2=-9^2\] \(\color{blue}{\text{End of Quote}}\) you didn't complete the square
I'll show you what I did. Give me a moment to type it all
okay
\[x^2+y^2-6x+2y=-9 --> (x^2-6x+9)+(y^2+2y+1)=-9+9+1\] \[(x-3)^2+(y+1)^2=1\] (3,-1)
...I think I got the answer but I'm unsure. Sorry I've been meaning to reply for awhile but uh.. I probably got it wrong but this is what I got from it: Center of the circle is: C = (-6, 2) Radius of the circle is 9.
...I think I got the answer but I'm unsure. Sorry I've been meaning to reply for awhile but uh.. I probably got it wrong but this is what I got from it: Center of the circle is: C = (-6, 2) Radius of the circle is 9.
yes your work is correct remember to complete the square we should we should divide b by 2 and then add (b/2)^2 to the opposite side that's how u got +9 and 1 at right side -9+9+1= 1 so yes your work is correct good job!
ok but @peridot got a different answer :/
\[x^2+y^2-6x+2y=-9 --> (x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+9+1\] how did you get 9 ? explain
-6 = -3, -3^2 =9
-6=-3 ? is there a typo ?
I thought you had to take half of the middle number, and then square it
oh yes so -6/2 = -3 -6 isn't equal to -3 alright and yes (-3)^2 =9 and like i said you should (b/2)^2 to the right side \[x^2+y^2-6x+2y=-9 --> (x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+\color{ReD}{9}+1\]
and same thing with one you take half of the middle term 2/2 =1 and then take square add to the right side \[x^2+y^2-6x+2y=-9 --> (x^2-6x+\color{Red}{9})+(y^2+2y+1)=-9+\color{ReD}{9}+\color{blue}{1}\] -9+9 +1 = 1 so if your work is correct then obviously your answer is correct
ok but I need to find out what the radius is now
\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia \[(x-h)^2+(y-k)^2=r^2\] ? \(\color{blue}{\text{End of Quote}}\) general form of the circle write where r is the radius
now compare your equation with that general form what is the radius in that equation ?
So radius is 1?
going to walmart, can i just please have answer? this is confusing me more than it should :(
\(\color{blue}{\text{Originally Posted by}}\) @rawrolivia So radius is 1? \(\color{blue}{\text{End of Quote}}\) yes right
you already know the answer..
\[(x-h)^2+(y-k)^2=r^2\] general form of the circle write where r is the radius and (h,k) is the center
I went to school to attend a session. Sorry. Let me know if you want to further learn or go over about circles.
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