Question 1107302
rewrite as x^2-4x+y^2-6y=-9
complete the square for x and y
x^2-4x+4+y^2-6y+9=-9+13
(x-2)^2+(y-3)^2=4
This is a circle with center (2, 3) and radius 2
It touches the y-axis where x=0
Therefore, (-2)^2+(y-3)^2=4
(y-3)^2=0
y=3, so it touches at (0,3) and is a "bounce," 3 units from the origin.
{{{graph(300,300,-10,10,-10,10,3-sqrt(-(x-2)^2+4),3+ sqrt(-(x-2)^2+4))}}}