Question

Graph each circle by rewriting the equations in standard form. X^2+y^2+4x-4y-1=0

Answer 1

`\begin{gathered} x^2+y^2+4x-4y-1=0 \\ \text{group them in terms of }x\text{ and }y \\ \mleft(x^2+4x\mright)+(y^2-4y)=1 \\ \text{complete the squares in terms of }x\text{ and }y \\ \mleft(x+2\mright)^2-2^2+(y-2^2)-2^2=1 \\ \mleft(x+2\mright)^2-4+(y-2)^2-4=1 \\ \text{Simplify} \\ (x+2)^2+(y-2)^2=1+4+4 \\ (x+2)^2+(y-2)^2=9\Rightarrow(x+2)^2+(y-2)^2=3^2 \\ \text{If we compare this to the equation of a circle} \\ (x-h)^2+(y-k)^2=r^2 \\ \text{Then it means that }(x+2)^2+(y-2)^2=3^2,\text{ is a circle with }(-2,2)\text{ as the center}_{} \\ \text{with a radius of 3} \end{gathered}`