We have,
Equation of circle
x2+y2−4x+6y−12=0
Comparing that,
x2+y2+2gx+2fy+c=0
g=−2,f=3,c=−12
Then,
Radius of circle
=√g2+f2−c
=√(−2)2+(3)2−(−12)
=√4+9+12
=√25
=5
If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is: