equation of a circle and it's area

We can complete the squares of the circle's equation:

$\begin{aligned} \bigg(x^2 - 6y + {\color{#20A900}{(9)}} \bigg) + \bigg( y^2 -8y + {\color{#20A900}{(16)}} \bigg) - 16 &= {\color{#20A900}{(9)}} + {\color{#20A900}{(16)}} \\ (x-3)^2 + (y-4)^2 &= {\color{#20A900}{(41)}} \\ r^2 &= 41 \end{aligned}$

$\text{Area } = \pi r^2 \implies \boxed{41 \pi}$

Equation of the circle is $(x-3)^2+(y-4)^2=41$ , Hence it's radius is $\sqrt{41}$ .

Therefore area of circle is given by $\pi(\sqrt{41})^2=\boxed{41\pi}$ .

Radius of circle with equation

$x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$

Putting g = -3, f = -4 and c = -16, we get $\boxed{r = \sqrt{41}}$

So, Area = $\pi r^2$

$\boxed{Area = 41\pi}$