Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions?

Question

Question Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions

My attempt:

If we try to solve this problem by substituiton the second equations into the first we get:

$$x^2+4(x-2)=k$$

which has discriminant $48+4k$. Thus the equation has $0,1$ and $2$ solutions when $k<-12$, $k=-12$ and $k>-12$ respectively.

However I notice this doesn't match the picture where it's clear the nature of the solutions changes when $k=4$ , why is this?

Answer 1

As a function of $k$, the number of solutions to the equation $x^2+4(x-2)=k$ is not equal to the number of solutions to the original equation. When you substituted, you lost information.

It is true that if $x^2+4(x-2)=k$ has no solutions, then the original system has no solutions. But the converse is not true. The system might have no solutions even though $x^2+4(x-2)=k$ does. For example, with $k=-3$, we get solutions $x=1, -5$ to the substituted equation. But neither of these gives a solution for $y^2=4(x-2)$.

The real constraint on the number of solutions to the system is that $y^2\geq0$. When $y=0$, we get $x=2$, which by the other equation implies $k=4$.

Answer 2

k is non negative. you are looking for the intersection points of the circle with radius square root of k and a parabola opening to the right and cutting the x-axis at x=2 and symmetric about the x-axis. So if square root of k is less than 2 no solution and if it is 2, single solution and if greater than 2 then two solutions

Answer 3

The equation $y^2=4(x-2)$ implies that $x\geq2$. That means the curve defined by $y^2=4(x-2)$ (a parabola) has no point for $x<2$. So, if you are looking for its intersection with another curve (here $x^2+y^2=k$) you should look over $x>2$.

For example, if the goal is having no intersection you should find $k$ such that the equation$$x^2+4(x-2)=k$$has no roots over $x>2$. This is equivalent to find $k$ such that the greatest root of the equation happens before $x=2$. Thus by solving the quadratic equation we have $$x^*=\frac{1}{2}\left(-4+\sqrt{48+4k}\right)<2$$ which leads to$$k<4$$