If circle $x^2+y^2=4$ intersects the hyperbola $xy=4$ in for points $(x_i,y_i) : i=1,2,3,4$ then find $$\prod_{i=1}^4 x_i$$
But when I graph it, these do not intersect. So am I wrong or is the question itself wrong printed??
[Math] Circle and hyperbola intersection
circlesconic sections
Related Solutions
Your intuition can be stated as the proposition below:
If $A$ and $B$ are two different points on a hyperbola and the midpoint of $AB$ lies on the asymptotes, then this midpoint is the center of the hyperbola.
Proof: Note that this proposition is irrelevant to coordinate systems, thus without loss of generality, the hyperbola can be assumed to be $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Suppose the coordinates of $A$ and $B$ are $(x_1, y_1)$ and $(x_2, y_2)$, respectively, then $x_1, x_2 ≠ 0$ and$$ \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1, \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1. \tag{1} $$ Because the asymptotes are $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$ and $\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)$ lies on the asymptotes, then$$ \frac{(x_1 + x_2)^2}{a^2} - \frac{(y_1 + y_2)^2}{b^2} = 0. \tag{2} $$
Now, suppose $x_1 + x_2 ≠ 0$, then $y_1 + y_2 ≠ 0$ by (2). From (1) there is$$ \frac{x_1^2 - x_2^2}{a^2} = \frac{y_1^2 - y_2^2}{b^2}, $$ and from (2) there is$$ \frac{(x_1 + x_2)^2}{a^2} = \frac{(y_1 + y_2)^2}{b^2}, $$ thus\begin{align*} &\mathrel{\phantom{\Longrightarrow}}{} \frac{x_1 - x_2}{x_1 + x_2} = \frac{\dfrac{x_1^2 - x_2^2}{a^2}}{\dfrac{(x_1 + x_2)^2}{a^2}} = \frac{\dfrac{y_1^2 - y_2^2}{b^2}}{\dfrac{(y_1 + y_2)^2}{b^2}} = \frac{y_1 - y_2}{y_1 + y_2}\\ &\Longrightarrow \frac{2x_1}{x_1 + x_2} = \frac{x_1 - x_2}{x_1 + x_2} + 1 = \frac{y_1 - y_2}{y_1 + y_2} + 1 = \frac{2y_1}{y_1 + y_2}\\ &\Longrightarrow \frac{y_1}{x_1} = \frac{y_1 + y_2}{x_1 + x_2} \Longrightarrow \frac{y_1}{x_1} = \frac{y_2}{x_2} := c. \end{align*} Note that $x_1 + x_2 ≠ 0$, plugging $y_1 = cx_1$ and $y_2 = cx_2$ into (2) to get $\dfrac{1}{a^2} - \dfrac{c^2}{b^2} = 0$, then by (1) there is$$ 1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = \left( \frac{1}{a^2} - \frac{c^2}{b^2} \right) x_1^2 = 0, $$ a contradiction. Therefore, $x_1 + x_2 = 0$, which implies $y_1 + y_2 = 0$ by (2). Hence the midpoint of $AB$ is the center $(0, 0)$ of the hyperbola.
The line equation seems a bit of a distraction. Here's how I'd solve the problem.
Suppose $H$ and $K$ determine a chord of the hyperbola that subtends a right angle at its center (the origin), $O$. We can write $$H = (h \cos\phi, h \sin\phi) \qquad K = (k \sin\phi,-k \cos\phi)$$ where $h:=|OH|$, $k:=|OK|$, and $\phi$ is some arbitrary angle. Define $p := |OP|$, where $P$ is the foot of the altitude from $O$ to $\overleftrightarrow{HK}$. Calculating the area of $\triangle HOK$ in two ways, we have
$$\frac12|OH||OK| = \frac12|OP||HK|\quad\to\quad hk = p\sqrt{h^2+k^2}\quad\to\quad p = \frac{1}{\sqrt{\dfrac{1}{h^2}+\dfrac{1}{k^2}}} \tag{1}$$
Since $H$ and $K$ both lie on the hyperbola, we have $$\begin{align} \frac{h^2}{a^2}\cos^2\phi - \frac{h^2}{b^2}\sin^2\phi &= 1 \tag{2} \\[4pt] \frac{k^2}{a^2}\sin^2\phi - \frac{k^2}{b^2}\cos^2\phi &= 1 \tag{3} \end{align}$$ Therefore, $k^2 (2) + h^2 (3)$ becomes $$\frac{h^2k^2}{a^2}(\cos^2\phi+\sin^2\phi)-\frac{h^2k^2}{b^2}(\sin^2\phi+\cos^2\phi) = h^2 + k^2 \quad\to\quad \frac{1}{a^2}-\frac{1}{b^2} = \frac{1}{h^2} + \frac{1}{k^2} \tag{4} $$ Thus, recalling $(1)$, we have $$p = \frac{1}{\sqrt{\dfrac{1}{a^2}-\dfrac{1}{b^2}}} = \frac{ab}{\sqrt{b^2-a^2}} \tag{5}$$ Since $p$ is the distance from $\overleftrightarrow{HK}$ to the origin, we conclude that the line is always tangent to the circle with the radius given in $(5)$, as desired. $\square$
Best Answer
You are right.
Multiply the first equation by $x^2$.
$$x^4+x^2y^2=4x^2$$ and subsitute $x^2y^2$:
$$x^4+16=4x^2.$$
This biquadratic equation has no real solutions.
Anyway, if you consider the complex solutions, by the Vieta formulas, the product of the roots is just the constant term, $16$.