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 .
EP is right but maybe that does not help much if you do not know how to reduce the size.
Have you looked at tthe thread included below. Maybe one of the suggestions in there will work for you.
I have stated in there how I do it and other people have stated how they do it.
https://web2.0calc.com/questions/how-to-upload-a-picture_1
(this was just the 4th thread in the sticky topics by the way.)