# Circle and hyperbola intersection

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??

• Right, these curves do not intersect. However, there might be complex solutions.
– zoli
Jan 10, 2018 at 8:42
• @zoli: right, but considering the "complex" intersection points is a weird thing.
– user65203
Jan 10, 2018 at 8:47
• @YvesDaoust: Weird, indeed.
– zoli
Jan 10, 2018 at 8:56

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, $16$.

• Yes the answer is indeed 16, so could you elaborate please
– Sri
Jan 10, 2018 at 8:49
• @sri Lookup Vieta's formulas.
– user65203
Jan 10, 2018 at 8:50
• In the vieta formula for product of roots, what should be $a_0$ and $a_n$ in $product=(-1)^n \frac{a_0}{a_n}$
– Sri
Jan 10, 2018 at 9:02
• @Sri: I answered that. Spend some effort.
– user65203
Jan 10, 2018 at 10:17
• yes i did do that but i get -16. for circle --> $(-1)^2 \frac{4}{1} = 4$ and for hyperbola $(-1)^1 \frac{4}{1} = -4$. thus, $(-4)(4) = -16$. that is why i asked you. i do not understand how i should use this
– Sri
Jan 10, 2018 at 10:56

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.

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.

• Yeah there are no real intersection of the roots I know that..I have just formed 2 different equations ..what's.wrong with the solution ? The 2 equations in my solution obviously do not have any real intersection Jan 10, 2018 at 9:01

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:

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

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

The hyperbola and the circle have no common points.