The general form of the equation of a circle is given as:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
Where:
(a, b) is the coordinate of the center of the circle
r is the radius of the circle
Now, given the circle in question, with:
(-4, 12) as the coordinate of the center
sqrt(3) as the radius of the circle,
The equation of the circle is as follows:
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-(-4))^2+(y-12)^2=(\sqrt[]{3})^2 \\ (x+4)^2+(y-12)^2=(\sqrt[]{3})^2 \\ (x+4)^2+(y-12)^2=3^{} \end{gathered}[/tex]
Therefore, the equation of a circle with center (-4, 12) and radius sqrt(3) is:
(x + 4)^2 + (y - 12)^2 = 3
