Answer:
Given that the equation of a circle is :
[tex] \green{ \boxed{\boxed{\begin{array}{cc} {x}^{2} + {y}^{2} = 6x - 8y \\ = > {x}^{2} + {y}^{2} - 6x + 8y = 0 \\ = > {x}^{2} + {y}^{2} + 2 \times ( - 3) \times x + 2 \times 4 \times y = 0 \\ \\ \sf \: standard \: equation \: o f \: circle \: is : \\ {x}^{2} + {x}^{2} + 2gx + 2fy + c = 0 \\ \\ \sf \: by \: comparing \\ \\ g = - 3 \\ f = 4 \\ c = 0 \\ \\ \sf \: radius \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ = \sqrt{ {( - 3)}^{2} + {4}^{2} - 0 } \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5 \: unit \\ \\ \bf \: area \: = \pi {r}^{2} \\ = \pi \times {5}^{2} \\ =\pink{ 25\pi \: { unit }^{2} }\end{array}}}}[/tex]
