Question

Find the standard form Center and radius of the following circles:4x^2 + 4y^2 + 36y + 5 = 0

Answer 1

Circle equation:

(x-a)^2 + (y-b)^2 = r^2

x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2

x^2 + y^2 + a^2 + b^2 - 2ax - 2by - r^2 = 0
_______________________

4x^2 + 4y^2 + 36y + 5 = 0

4x^2 + 4y^2 + 36y + 5 +0x + 0y = 0

(-2ax = 0x)

-2a = 0

a = 0

(-2by = 0y)

-2b = 0

b = 0

____________________

And now:

a^2 + b^2 - r^2 = 5

0^2 + 0^2 - r^2 = 5

0+0-r^2 = 5

r^2 = -5

Since there is no negative radius, we say that it is not an equation of a circle