# 2010 AMC 12A Problems/Problem 13

## Problem

For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

## Solution

The image below shows the two curves for $k=4$. The blue curve is $x^2+y^2=k^2$, which is clearly a circle with radius $k$, and the red curve is a part of the curve $xy=k$.

$[asy] import graph; size(200); real f(real x) {return 4/x;}; real g1(real x) {return sqrt(4*4-x*x);}; real g2(real x) {return -sqrt(4*4-x*x);}; draw(graph(f,-20./3,-0.6),red); draw(graph(f,0.6,20./3),red); draw(graph(g1,-4,4),blue); draw(graph(g2,-4,4),blue); axes("x","y"); [/asy]$

In the special case $k=0$ the blue curve is just the point $(0,0)$, and as $0\cdot 0=0$, this point is on the red curve as well, hence they intersect.

The case $k<0$ is symmetric to $k>0$: the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$.

Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$.

At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution.

For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$. To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$. By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$, and that equality holds whenever $x^2 = (k/x)^2$, i.e., $x=\pm\sqrt k$.

Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$. Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$.

Clearly the only such integer is $k=1$, hence the two curves are only disjoint for $k=1$ and $k=-1$. This is a total of $\boxed{2\ \textbf{(C)}}$ values.

## Solution 2

From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.

Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates $(x, k/x)$ derived from the equation. Then, all coordinates that satisfy $\sqrt{ x^2+ (k/x)^2 } \leq k$ intersect the circle. Squaring, we find $x^2+(k/x)^2 \leq k^2.$ After multiplying through by $x^2$ and rearranging, we find $x^4-x^2k^2+k^2 \leq 0$. We see this is a quadratic in $x^2$ and consider taking the determinant, which tells us that solutions are real when, after factoring: $k^2(k^2-4) \leq 0$ We plot this inequality on the number line to find it is satisfied for all values except: $(-1, 0, 1)$. We then eliminate 0 because it is extraneous as both $xy=0$ and $x^2+y^2=0$ are points which coincide. Therefore, there are a total of $\boxed{2\ \textbf{(C)}}$ values.

## See also

 2010 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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