+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?
+9466 I think you’re pretty much on the right track 
Assuming: \(x^2 + y^2 - 6x + py + q = 0\)
\(x^2 + y^2 - 6x + py + q = 0\)
Subtract q from both sides of the equation
\(x^2 + y^2 - 6x + py = -q\)
Rearrange the terms on the left side of the equation.
\(x^2 - 6x + y^2 + py = -q\)
Add 9 and add \((\frac{p}{2})^2\) to both sides of the equation.
\(x^2 - 6x + 9 + y^2 + py + (\frac{p}{2})^2 = -q + 9 + (\frac{p}{2})^2\)
Factor both perfect square trinomials on the left side.
\((x - 3)^2 + (y + \frac{p}{2})^2 = -q + 9 + (\frac{p}{2})^2\)
Now we can see that the center of the circle is the point (3, -\(\frac{p}{2}\))
Because the circle is tangent to the y-axis,
radius = distance between center and y-axis
radius = distance between (3, -\(\frac{p}{2}\)) and (0, -\(\frac{p}{2}\))
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 .
