Let p and q be constants such that the graph of x^2+y^2-6x+py+q=0 is tangent to the y-axis. What is the area of the region enclosed by the graph?

I have no idea what to do? Can someone please help me with the solution to the problem?

2 answers

x^2+y^2-6x+py+q=0

(x-3)^2 + (y+p/2)^2=-q-9-P^2/4
so radius of this circle is
sqrt(q-9-p^2/4)

area= PI (q-9-p^2/4)

check my thinking.
For the area, wouldn't be (q-9-p^2/4)^2 because the formula for the circle is pi radius squared?