[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
As we know, the standard form of circle is written in this way ~
[tex] \qquad \sf x{}^{2} + y {}^{2} + 2gx + 2fy + c = 0[/tex]
where, the coordinates of centre is (-g , -f) and radius equals to :
[tex]\qquad \sf \dashrightarrow \: \sqrt{g {}^{2} + {f}^{2} - c } [/tex]
Now, it's time to equate the coordinates of centre ~
- [tex] \qquad \sf( - g, - f) \: \: and \: \: (0,4)[/tex]
Here we get,
[tex]\qquad \sf \dashrightarrow \: - g = 0[/tex]
[tex]\qquad \sf \dashrightarrow \: g = 0[/tex]
and
[tex]\qquad \sf \dashrightarrow \: - f = 4[/tex]
[tex]\qquad \sf \dashrightarrow \: f = - 4[/tex]
Now, let's find the the Radius using distance formula on the given points, one of them is centre and other is point lying on circle, so distance between them is the radius.
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{( 0 - ( - 2)) {}^{2} + (4 - ( - 4)) {}^{2} }[/tex]
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{ (0 + 2) {}^{2} + (4 + 4) {}^{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{ (2) {}^{2} + (8) {}^{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{ 4+ 64 } [/tex]
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{ 68 } [/tex]
[tex]\qquad \sf \dashrightarrow \: r = 2\sqrt{ 17 } [/tex]
Now, use the the following equation to find c of the standard equation
[tex]\qquad \sf \dashrightarrow \: r = \sqrt{g {}^{2} + {f}^{2} - c} [/tex]
[tex]\qquad \sf \dashrightarrow \: 2 \sqrt{17} = \sqrt{ {0}^{2} + {4}^{2} - c}[/tex]
squaring both sides :
[tex]\qquad \sf \dashrightarrow \: 68 = {0}^{2} + {4}^{2} - c[/tex]
[tex]\qquad \sf \dashrightarrow \: 68 = 8 - c[/tex]
[tex]\qquad \sf \dashrightarrow \: - c = 68 - 8[/tex]
[tex]\qquad \sf \dashrightarrow \: c = - 60[/tex]
Therefore, we got the standard equation of circle as ~
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + {y}^{2} + 2(0)x + 2( - 4)y + ( - 60) = 0[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + {y}^{2} - 8y - 60 = 0[/tex]
I Hope it helps ~
