Question

Determine which of the following equations represents a circle with a real non-zero radius: (d) 360^2x +36y^2 - 36x +48y = -16

Answer 1

We have the following:

`36x^2+36y^2-36x+48y=-16`

The equation of a circle when the center is not the origin is as follows

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (h,k)\rightarrow center \\ r\rightarrow radius \end{gathered}`

we organize and solve

`\begin{gathered} (36)/(36)x^2+(36)/(36)y^2-(36)/(36)x+(48)/(36)y=-(16)/(36) \\ x^2+y^2-x+(4)/(3)y=-(4)/(9) \\ (x^2-x+(1)/(4))+(y^2+(4)/(3)y+(4)/(9))=-(16)/(36)+(1)/(4)+(4)/(9) \\ (x-(1)/(2))^2+(y+(2)/(3))^2=(1)/(4) \\ (x-(1)/(2))^2+(y+(2)/(3))^2=((1)/(2))^2 \\ \text{center}=((1)/(2),-(2)/(3)) \\ \text{radius}=(1)/(2) \end{gathered}`

Therefore, in this case if we have a circle with center at (1/2, -2/3) and the one with a radius of 1/2