Question

1. find the equation of the circle with center at (-3,1) and through the point (2,13)

Answer 1

Try this solution:
Common view of the equation of the circle is (x-a)²+(y-b)²=r², where point (a;b) is centre of the circle, r - radius.
1. using the coordinates of the centre and point (2;13) it is possible to define the radius of the circle: r=√(5²+12²)=13;
the equation is (x+3)²+(y-1)²=13² or (x+3)²+(y-1)²=169;
2. using the coordinates of the centre and the radius: (x-2)²+(y-4)²=6² or (x-2)²+(y-4)²=36.