Let C be the circle in R² having the point (h, k) and (0, 1) as diameter. 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.
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
$\displaystyle \left(\frac{h}{2},\frac{k+1}{2}\right)$
Radius of Circle
$\displaystyle \sqrt{(h-0)^{2}+(k-1)^{2}}$
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 $\displaystyle \left(\frac{h+0}{2}, \frac{k+1}{2}\right)= \left(\frac{h}{2}, \frac{k+1}{2}\right)$ 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 $\displaystyle \left(\frac{h}{2}, \frac{k+1}{2}\right)$ is the center and (0, 0) is on the circumference, the radius is $\displaystyle \sqrt{\frac{h^2}{4}+ \frac{(k+1)^2}{4}}$
The equation of a circle with center at (a, b) and radius r is $\displaystyle (x- a)^2+ (y- b)^2= r^2$ so the equation of this circle is $\displaystyle \left(x- \frac{h}{2}\right)^2+ \left(y- \frac{k+1}{2}\right)^2= \frac{h^2+ (k+1)^2}{4}$.
The first thing we can do is multiply both sides of that by 4, leaving $\displaystyle (2x- h)^2+ (2y- (k+1))^2= h^2+ (k+1)^2$.
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, $\displaystyle ax^2+ bx+ c= 0$ is $\displaystyle b^2- 4ac$.)