If two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then

the circle $${x^2} + {y^2} + 4x - 7y + 12 = 0$$ cuts an intercept on Y-axis of length

$$S\equiv x^2+y^2+2x+3y+1=0$$ and $$S'\equiv x^2+y^2+4x+3y+2=0$$ are two circles. The point $$(-3,-2)$$ lies

What will be the equation of circle whose centre is (1, 2) and touches X-axis?