a circle has an equation 2(x-3)^2+2y^2=8 I need to find center, radius and intercepts.
3,0 and rad sqrt(8) I think it might not be a circle tho.
general equation of a circkle (x-h)^2+(y-k)^2=r^2 where h,k is center
I am totally confused on how to set up these equations for circles.
Radius should be 2; if the equation is 2(x-3)^2+2y^2=8 factor out 2 and we have; 2[(x-3)^2+y^2]=8, (x-3)^2+y^2=4, so radius is sqrt(4)=2
Ok what's the problem?
where did you get 4?
If we factor out 2,we can divide both members by 2
oh ok I see...
Do you still want x intercepts?
yes
Ok we have (x-3)^2+y^2=4, if we want x-iintercepts we use y=0, so we have (x-3)^2=4; x^2-6x+9=4; x^2-6x+5=0; x=(6+-sqrt(36-4*5))/2; x=(6+-4)/2 ; x=3+-2; x=1 and x=5
That's are x-intercepts, you draw that points in a paper, and beacause it's a circle you see that there aren't y-intercepts
Thank you!! So let me see if this is correct. for another problem, I m working on.. .(x-5)^2+y^2=100 Could I do: (x-5)^2+y^2=5? and my answer would be sqrt(5)=2?
No, we can factor out 2 above because it's multipliying x terms and y terms, in (x-5)^2+y^2=100 we can't factoru out nothing so the equation is in it's simplist form so radius is sqrt(100)=10
its*
that is what I had first, so since the number is ^2 we have to use that number to factor out.
No haha, ok imagine we have this equation 2*x^2+2*y^2=4; it's the same that 2*x*x+2*y*y=4 right?
I would say yes.. This whole math this is so confusing and frusrating to me..lol
Ok don't worry it's just practice x^2=x*x , x^3=x*x*x that's what that symbol means ok?
ok... Thanks for your help..
yw, so this equation 2*x^2+2*y^2=4 is the same that 2*x*x+2*y*y=4 rigth?
If we divide both members by 2, we have x*x+y*y=2 right?
It's the same we did 2(x-3)^2+2y^2=8
here*
Do you understand now?
yes.. thank you again
np
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