Question

Find the equation of the circle that passes through the origin and has its center at (-3,-4).

Answer 1

$equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)\ is\ a\ cener\ of\ the\ circle\\\\(a;\ b)=(-3;-4)\\\\r-radius\ of\ the\ circle\\\\r=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\\\Answer:(x+3)^2+(y+4)^2=5^2\to(x+3)^2+(y+4)^2=25$

Answer 2

If the center is at (-3, -4) and the origin is on the circle, then the radius
of the circle is the distance between the origin and (-3, -4).

R² = (-3)² + (-4)²
R² =  9    +  16
R² =      25
R = 5

The equation of the circle is [  (x + 3)² + (y + 4)² = 25 ].