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The problem is this:

to show that a circle of radius $r$ and center $(h, k)$ intersects the $x$-axis at exactly one point if and only if $\left|k\right| = r$.

Using geometrical intuition, this is easy to see. When a circle's bottom or top is just touching the $x$-axis, its center is one radius away. (Note that I put "tangent" in quotes because I'm using a definition which captures one aspect of tangency, but doesn't hold in the case of, say, intersecting lines.)

I'm having some trouble with the forward direction of this statement.

My attempt so far

Suppose that the circle $C$ touches the $x$-axis at exactly one point. That is, suppose that $(\alpha, 0) \in C$ for some $\alpha \in \mathbb{R}$ and $(x, 0) \notin C$ for all $x \neq \alpha$ ($x \in \mathbb{R}$). Then

$$ \begin{align*} (\alpha - h)^2 + (0 - k)^2 &= r^2 \\ \Rightarrow (\alpha - h)^2 + k^2 &= r^2. \end{align*} $$

Somehow I need to use the fact that the only $\alpha$ which works is $h$... but how can I show that $\alpha = h$?

The backward direction

Suppose $\left|k\right| = r$. Then

$$ \begin{align*} (h - h)^2 + (0 - k)^2 &= k^2 \\ &= \left|k\right|^2 \\ &= r^2 \\ \Rightarrow (h, 0) &\in C \end{align*} $$

And for other $x \neq h$, we have $(x - h)^2 + (0 - k)^2 = (x - h)^2 + r^2 > r^2$.

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Let us find the intersection of the circle with the $x$-axis ($y=0$).

Setting $y=0$ in the equation $$(x-h)^2+(y-k)^2=r^2,$$ we get $$x^2-2hx+h^2+k^2-r^2=0.$$

Each root represents the abscissa of one intersection.

For tangency, the coordinate of two intersections must coincide $\implies $ the roots of must be the same.

For the problem in the question, we have $(\alpha - h)^2 + (0 - k)^2 = r \Rightarrow \alpha = h \pm \sqrt{r^2 - k^2}$. Then, since it intersects the $x$-axis at only one point, we must have $\sqrt{r^2 - k^2} = 0$, i.e. $r^2 = k^2$ or $r = \left|k\right|$.

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  • $\begingroup$ What do you mean by "For tangency the coordinate of two intersections must coincide ⟹ the roots of must be same"? $\endgroup$ – AmadeusDrZaius Jul 2 '14 at 16:57
  • $\begingroup$ @AmadeusDrZaius, the roots of the quadratic equation will have to be same for the tangency $\endgroup$ – lab bhattacharjee Jul 2 '14 at 16:58
  • $\begingroup$ Ah, I get it now. $\endgroup$ – AmadeusDrZaius Jul 2 '14 at 17:01
  • $\begingroup$ @AmadeusDrZaius, Find my other answer, too:) $\endgroup$ – lab bhattacharjee Jul 2 '14 at 18:30
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The distance of the tangent from the center = the radius

$$\frac{|k|}{\sqrt{0^2+1^2}}=r$$

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You've got a quadratic equation in $\alpha$ in terms of known constants $h$,$k$,$r$. So if circle touches x-axis, then what should you expect the solution to your quadratic equation to be? (in terms of its multipilicty). And then think about discriminant

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Geometrically, you know that the radius of the circle is perpendicular to the tangent; since the tangent is horizontal, the radius to the intersection point must be vertical...

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