Write the equation x^2+y^2-2x-4y-11=0 in standard form.
x^2-2x+1+y^2 -4y+4-11-1+4=0
(x-1)^2 + (y-2)^2=8
Wait! It just appears (x-1)^2 + (y-2)^2=16 and (x-1)^2 + (y-2)^2=4 so it would be this one?
I think its this one (x-1)^2 + (y-2)^2=16
what about you?
To write the equation x^2+y^2-2x-4y-11=0 in standard form, we need to complete the square for both x and y terms.
First, let's rearrange the equation by grouping the x and y terms together:
x^2 - 2x + y^2 - 4y - 11 = 0
To complete the square for x terms, we want to add and subtract the square of half the coefficient of x (which is -1 in this case):
(x^2 - 2x + (-1)^2) + (y^2 - 4y - 11) = 1 + (-11)
This can be simplified to:
(x - 1)^2 + (y^2 - 4y - 11) = -10
Similarly, let's complete the square for y terms:
(x - 1)^2 + (y^2 - 4y + (-2)^2 - (-2)^2 - 11) = -10 + 4
This can be simplified to:
(x - 1)^2 + (y - 2)^2 - 15 = -6
Finally, let's move the constant term to the other side of the equation to put it in standard form:
(x - 1)^2 + (y - 2)^2 = -6 + 15
This simplifies to:
(x - 1)^2 + (y - 2)^2 = 9
Therefore, the equation x^2+y^2-2x-4y-11=0 in standard form is (x - 1)^2 + (y - 2)^2 = 9.