+6 Let p and q be constants such that the graph of x^2 + y^2 - 6x +py + q is tangent to the y-axis. What is the area of the region enclosed by the graph?
So I have moved it around to be the standard equation to finding the radius of a circle (x-h)^2+(y-k)^2=r^2
And i ended up with (x-3)^2+(y+(p/2))^2=-q-9-(p/2)^2
I have the feeling that I'm over complicating this could anyone give me some help?
+9488 I think you’re pretty much on the right track 
Assuming: x2+y2−6x+py+q=0
x2+y2−6x+py+q=0
Subtract q from both sides of the equation
x2+y2−6x+py=−q
Rearrange the terms on the left side of the equation.
x2−6x+y2+py=−q
Add 9 and add (p2)2 to both sides of the equation.
x2−6x+9+y2+py+(p2)2=−q+9+(p2)2
Factor both perfect square trinomials on the left side.
(x−3)2+(y+p2)2=−q+9+(p2)2
Now we can see that the center of the circle is the point (3, -p2)
Because the circle is tangent to the y-axis,
radius = distance between center and y-axis
radius = distance between (3, -p2) and (0, -p2)
radius = 3
area = π · radius2
area = π · 32
area = 9π sq. units
You can play around with the sliders on this graph to check:
https://www.desmos.com/calculator/zvm1f8xhaf
Notice that the distance between the center and the y-axis stays 3 .
