Thread: circle

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



Radius of Circle



Where's the tricky part?

i cant even understand the question is that the question says that find the radius/

Originally Posted by apple2009

Let C be the circle in R² having the point (h, k) and (0, 1) as diameter.

As TKHunny says, two "points" do not form a diameter- you mean the line segment having those points as endpoints.

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 center of a circle is the midpoint of any diameter so this circle has as center.

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 .)

thank you very much

There is a circle equation formed from 2 end points of a diameter but it does not seem to be regularly taught.