Question

What is the equation of a circle with its center at (0,-2) and passing through the point (3,-5)

Answer 1

Circle's Standard Form Equation:
$r^{2} = (x - xo)^{2} + (y - yo)^{2}$

First you need to substitute the C (0,-2) in the equation, where: Xo = 0 and Yo = -2;

$r^{2} = (x - 0)^{2} + (y - ( -2))^{2} \\ r^{2} = x^{2} + (y + 2)^{2}$

Now you substitute the Point (3,-5) values in the last expression, where: X = 3 and Y = -5;

$r^{2} = 3^{2} + (( - 5) + 2)^{2} \\ r ^{2} = 9 + ( - 3)^{2} \\ r^{2} = 9 + 9 \\ {r}^{2} = 18$

Now you get the R^2 value and substitute in the equation that you found putting the C values on it:

$r^{2} = x^{2} + (y + 2)^{2} \\ x^{2} + (y + 2)^{2} = 18$

Here is your standard form equation of the circle.