First, it doesn't make sense. I am guessing the diameter is the line segment between (h,k) and (0,1)
Second, isn't this just an algebra problem?
Center of Circle
Radius of Circle
Where's the tricky part?
As TKHunny says, two "points" do not form a diameter- you mean the line segment having those points as endpoints.
The center of a circle is the midpoint of any diameter so this circle has as center.Prove that this circle intersects the x-axis if and only if h²-4k≥0 and in this case the two intercepts are the roots of the equation x²-hx+k=0.
The radius of a circle is the distance from the center to any point on the circumference. Since is the center and (0, 0) is on the circumference, the radius is
The equation of a circle with center at (a, b) and radius r is so the equation of this circle is .
The first thing we can do is multiply both sides of that by 4, leaving .
The circle "intersects the x-axis" where y= 0. Put y= 0 into that equation and solve for x. That will be a quadratic equation so it will have two distinct solutions if and only its discrimant is 0. (The "discriminant" of the quadrtic equation, is .)