The equation of a circle having centre at (h,ok) and with a radius 'r' is (x-h)^2 + (y-ok)^2 = r^2 substituting the values (x +3)^2 + (y-4)^2 = 5^2 increasing, x^2 + 9 + 6x + y^2 + sixteen - 8y = 25 Simplifying, x^2 + y^2 + 6x -8y =0 examine: through fact the circle passes thro' foundation (0,0) substituting those values in the above ,we get 0 as answer.

X^2 + Y^2 -2X -4 = 0

x^2 + y^2 -2x - 4 = 0

(x-1)^2+(y-2)^2= 5 and (x-1)^2+(y+2)^2= 5

i.e., x^2+y^2-2x-4y=0 and x^2+y^2-2x+4y=0.