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Question:-

Find the interval of mm of each case, where there are 0, 1, 2, 3, or 4 solutions to the following system equation. (x and y are real-valued.)
$\begin{cases} & |x-1|+|y+1|=1\\&x^{2}+y^{2}=m^{2}\end{cases}$



$\large\text{\underline{Note:-{}}}$

Apply coordinate geometry

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Answer 1

$\large\text{\underline{Question}}$

Determine the interval of $m$ according to the number of solutions of a system equation $\begin{cases} & |x-1|+|y+1|=1 \\ & x^{2}+y^{2}=m^{2} \end{cases}$.

The number of solutions is 0, 1, 2, 3, or 4.

 

$\large\text{\underline{Footnote}}$

The first equation $|x-1|+|y+1|=1$ represents a circle of the taxicab geometry.  

The second equation $x^{2}+y^{2}=m^{2}$ represents a circle on the cartesian plane. So basically, we are solving the system equation of two circles in different branches of planes.

$\large\text{\underline{Solution}}$

In the taxicab geometry, the distance of two points is determined by the sum of vertical and horizontal distance.

The center of the circle is located at $(1,-1)$ and the radius is 1.

The four vertices of the taxicab circle are $A(1,0),\:B(2,-1),\:C(1,-2),\:D(0,-1)$, so we note that the circle is symmetric against a line $y=-x$.

Meanwhile, in the cartesian geometry, the circle $x^{2}+y^{2}=m^{2}$ represents a circle centered at $(0,0)$ and of the radius of $|m|$.

$\large\text{\underline{Footnote}}$

The number of intersections of a line and a circle can be 2, 1, or 0.

• 2 intersections if the line intersects the circle
• 1 intersection if the line is tangent to the circle
• 0 intersection if the line doesn't meet the circle

Let the midpoints of $\overline{DA}$ and $\overline{BC}$ be $M_{1}$ and $M_{2}$.

$\implies M_{1}(\dfrac{1}{2},\:-\dfrac{1}{2})$

$\implies M_{2}(\dfrac{3}{2},\:-\dfrac{3}{2})$

 

$\green{\text{Back to the solution,}}$

joining $\overline{OM_{1}}$ and $\overline{OM_{2}}$, the distances are $\dfrac{\sqrt{2}}{2}$ and $\dfrac{2\sqrt{2}}{3}$.

The intervals according to the graph;

• $0\leq|m|<\dfrac{\sqrt{2}}{2}$ then there is no solution. (0)
• $|m|=\dfrac{\sqrt{2}}{2}$ then there is one solution. (1)
• $\dfrac{\sqrt{2}}{2}<|m|<1$ then there are two solutions. (2)
• $1\leq |m|<\dfrac{3\sqrt{2}}{2}$ then there are two solutions. (2)
• $|m|=\dfrac{3\sqrt{2}}{2}$ then there are three solutions. (3)
• $\dfrac{3\sqrt{2}}{2}<|m|<\sqrt{5}$ then there are four solutions. (4)
• $|m|=\sqrt{5}$ then there are two solutions. (2)
• $|m|>\sqrt{5}$ then there is no solution. (0)

$\large\text{\underline{Conclusion}}$

0 solution if $-\dfrac{\sqrt{2}}{2}<m<\dfrac{\sqrt{2}}{2}$ or $m<-\sqrt{5}$ or $m>\sqrt{5}$.

1 solution if $m=\pm\dfrac{\sqrt{2}}{2}$.

2 solutions if $-\dfrac{3\sqrt{2}}{2}<m<-\dfrac{\sqrt{2}}{2}$ or $\dfrac{\sqrt{2}}{2}<m<\dfrac{3\sqrt{2}}{2}$ or $m=\pm\sqrt{5}$.

3 solutions if $m=\pm\dfrac{3\sqrt{2}}{2}$.

4 solutions if $-\sqrt{5}<m<-\dfrac{3\sqrt{2}}{2}$ or $\dfrac{3\sqrt{2}}{2}<m<\sqrt{5}$.