Question 890591
Something in your problem is unclear or was left out, so we cannot solve it without understanding what was meant by the problem.
I will make my best guess at what the problem meant, and I will solve that guessed problem. (No guarantees).
 
In general, you need 3 points to determine a circle,
so if we knew 2 different points where the circle crosses the x-axis,
we would only need another (third) point.
However, "touches x axis at (4,0)"
gives us point P(4,0) and strongly suggests the extra information that the x-axis is tangent to the circle
(that the circle touches the x-axis only at that point),
so that information is worth the same as 2 different points.
What I can only guess is what is meant by saying that
the circle "cuts up an intercept 8 units from the positive y axis"
 
The radius to the point of where a circle is tangent to a line is perpendicular to the line,
so we know that the radius is the vertical line {{{x=4}}} .
From that we have that the center is at {{{C(4,a)}}} , where {{{a}}} is the radius,
but we do not know {{{a}}} yet:
{{{drawing(300,300,0.1,0.9,-.1,0.7,
grid(0),circle(0.4,0.5,0.5),
green(line(0.4,-0.1,0.4,0.7)),
locate(0.41,-0.01,P(4,0)),
circle(0.4,0,0.01),
circle(0.4,0.5,0.01),
locate(0.41,0.49,C(4,a))
)}}}
 
Maybe what was meant is that the circle cuts a segment of length 8,
between its two intercepts, out of the positive y-axis, like this:
{{{drawing(300,300,-0.1,0.55,-0.05,0.6,
grid(0),circle(0.2,0.2sqrt(2),0.2sqrt(2)),
green(line(0.2,-0.2,0.2,0.8)),
locate(0.21,0.05,P(4,0)),
circle(0.2,0,0.01),
circle(0.2,0.2sqrt(2),0.01),
locate(0.21,0.28,C(4,a)),
blue(triangle(0,0.2sqrt(2)-0.2,0,0.2sqrt(2)+0.2,0.2,0.2sqrt(2))),
locate(0.01,0.3,blue(8)),locate(0.1,0.37,blue(a)),locate(0.1,0.23,blue(a))
)}}} and the blue isosceles triangle is key to find {{{a}}} .
Drawing a horizontal line to the center of the circle, I split into two right triangles ("coincidentally" also isosceles)
{{{drawing(300,300,-0.1,0.55,-0.05,0.6,
grid(0),circle(0.2,0.2sqrt(2),0.2sqrt(2)),
green(line(0.2,-0.2,0.2,0.8)),
locate(0.21,0.05,P(4,0)),
circle(0.2,0,0.01),
circle(0.2,0.2sqrt(2),0.01),
locate(0.21,0.28,C(4,a)),
blue(triangle(0,0.2sqrt(2),0,0.2sqrt(2)+0.2,0.2,0.2sqrt(2))),
blue(triangle(0,0.2sqrt(2)-0.2,0,0.2sqrt(2),0.2,0.2sqrt(2))),
blue(rectangle(0,0.2sqrt(2),0.02,0.3)),
locate(-0.02,0.3,blue(8)),locate(0.1,0.37,blue(a)),locate(0.1,0.23,blue(a)),
locate(0.01,0.38,blue(4)),locate(0.01,0.22,blue(4)),locate(0.08,0.28,blue(4)),
locate(0.1,0.37,blue(a))
)}}} Then we apply the Pythagorean theorem to get {{{a^2=4^2+4^2}}} .
From there we get to {{{a^2=16+16}}} ---> {{{highlight(a^2=32)}}} ---> {{{a=sqrt(32)}}} ---> {{{highlight(a=4sqrt(2))}}}
and then, knowing the coordinates of the center and the radius, we can write the equation of the circle:
{{{highlight((x-4)^2+(y-4sqrt(2))^2=32)}}}