Answer:
[tex]\large\boxed{(x-2)^2+(y+5)^2=148}[/tex]
Step-by-step explanation:
The standard form of an equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have the center (2, -5) → h = 2 and k = -5.
Therefore the equation is:
[tex](x-2)^2+(y-(-5))^2=r^2\\\\(x-2)^2+(y+5)^2=r^2[/tex]
The circle passes through the point (4, 7). Put x = 4 and y = 7 to the equation:
[tex](4-2)^2+(7+5)^2=r^2\\\\2^2+12^2=r^2\\\\4+144=r^2\\\\r^2=148[/tex]
Finally we have:
[tex](x-2)^2+(y+5)^2=148[/tex]
