How would you put 16 + 6x – x^2 – y^2 = 0 into graphing form?

Question

I know it's a circle and that the graphing form is (x – h)^2 + (y – k)^2 = r^2. Can someone help me?

Answer 1

Yes I think it is a circle also. What you can do is set x = 0 then solve for y = -4 or +4. Then you can set y = 0 and solve for x. -x^2 + 6x + 16 = 0; (-x + 8) (x + 2) so x = -2 or +8. Plot these points and you will start to have some what of an idea what you are dealing with.

Answer 2

The way I would do it is this:
Basically define the equation by Y.
E.g. Y^2 = 16 + 6X -X^2
Y = sqroot 16+6X -X^2
Y= sqroot of (8-X) (2+X)

Now just enter values for X and solve for Y
For example, when X= 0, then Y = 4
When Y= 0, then the equation becomes
0 = (8-X) (2+X). Divide both sides by either 8-x or 2+X. Left side becomes zero and you get X = -2 or 8.

So now you can plot your points:
Y= 0, X = -2 or 8
X= 0, Y = 4
And enter more values for X and solve for Y and plot these. For square root of decimals, use your calculator. You should get the top of a circle (an arc) not a full circle.

Answer 3

ok I can do this but I cant explain it to you so you will need to go through my work and figure out what I did (I wont skip any steps)

Given: 16+6x-x^2-y^2=0

(-1)(16+6x-x^2-y^2)=0(-1)

-16-6x+x^2+y^2=0

x^2-6x+y^2=16

now for the fun part

(x^2-6x+9)+y^2=16+9

(x-3)^2+y^2=25

(x-3)^2+y^2=5^2

I hope you can figure out what I did, not sure that I could explain it to you.

Answer 4

solve so y =. then put that in the calculator or any graphing utility.

Answer 5

just don't

Answer 6

idk