One point on the circle is (3,0) and the other is (6,6)
The center must be (3, b) since the circle is tangent to the x axis
Since the radial distance from (3,0) to (3, b) = the same radial distance from (6,6) to ( 3,b)
So....using the square of the distances we have that
b^2 = ( 6-3)^2 + (6 - b)^2
b^2 = 9 + 36 - 12b + b^2
12b = 45
b = 45/12 = 15/4 = the radius of the circle
So we equation of the circle is
(x - 3)^2 + (y - 15/4)^2 = (15/4)^2
Since (p,p) is on the circle we have that
(p - 3)^2 + ( p - 15/4)^2 =(15/4)^2
p^2 - 6p + 9 + p^2 - (15/2)p + (15/4)^2 = (15/4)^2
2p^2 - (6 + 15/2)p + 9 = 0
2p^2 - (27/2)p + 9 = 0
4p^2 - 27p + 18 = 0 factor as
(4p - 3) ( p - 6) = 0
Setting both factors to 0 and so;ving for p we get that p = 6 (we already know this)
or
p = (3/4)
So (p,p) = (3/4 , 3/4)
See the graph here : https://www.desmos.com/calculator/hasagqw5bw