Circle and hyperbola intersection

Question

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??

Answer 1

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.

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Anyway, if you consider the complex solutions, by the Vieta formulas, the product of the roots is just the constant term, $16$.

Answer 2

You are right, these two doesn't intersect. Indeed, if we suppose for a contradiction that they intersect in real plane, the equation $x^2+\frac{16}{x^2}=4$ must have real solutions. But that is $$x^4-4x^2+16 = 0 \implies_{a = x^2}\ a^2-4a+16 = 0$$ whose discriminant is $4^2-4\cdot16 < 0$ so it has no real solution as required. But there might be some complex solutions as stated.

Answer 3

replace y=4/x in the equation of circle and get two equations.. X^2 - (2i)x -4 =0 and X^2 +(2i)x -4=0

All you need is (product of roots of first equation) x ( product of roots of second equation) which is 16.

Answer 4

Look at the figure below:

They obviously have no intersection.

generally the circle $x^2+y^2=r^2$ and $xy=a$ have the following intersections:

(1) $r<\sqrt{2a}$

They don't intersect at all.

(2) $r=\sqrt{2a}$

They intersect and touch in points $(\sqrt a,\sqrt a)$

(3) $r>\sqrt{2a}$

They intersect in 4 points (x,y) where:

$$x=\pm\sqrt{{{r^2\pm\sqrt{r^4-4a^2}}\over{2}}}$$

Answer 5

$$-4=4-8=x^2+y^2-2xy=(x-y)^2,$$ which is impossible.

The hyperbola and the circle have no common points.