Algebra Problem No.4

Similar solution as @Aryan Sanghi 's

Note that $(x-2)^2 + y^2 = m$ is a circle with center at $(2,0)$ and radius $\sqrt m$ and $(y-2)^2 + x^2 = m$ is a circle with center at $(0,2)$ and radius $\sqrt m$ . The two circles have the same radius $\sqrt m$ . The intersecting points of the two circles are the roots of the system of the two equations. The system has a single root when the circles touch each other. From the figure we can see that the two circle touch each other with $2\sqrt m = 2\sqrt 2 \implies m = \boxed 2$ .

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This is equivalent to showing that circles $S_1 = (x-2)^2 + y^2 - m = 0$ and $S_2 = (y-2)^2 + x^2 - m = 0$ touch each other at one point

$\text{For circle } S_1, C_1 = (2, 0) \text{ and } r_1 = \sqrt{m}$

$\text{For circle } S_2, C_2 = (0, 2) \text{ and } r_2 = \sqrt{m}$

As they touch each other, so

$C_1C_2 = r_1 + r_2$ $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{m} + \sqrt{m}$ $2\sqrt{2} = 2\sqrt{m}$

$\color{#3D99F6}{\boxed{m = 2}}$

Assume that the solution is $(x_{0},y_{0})$ . To satisfy the given equation system, $x_{0} = y_{0}$ (because the system is symmetrical, means that if you replace one variable by another one, the system remains unchanged).

$\implies (x_{0}-2)^{2} + x_{0}^{2} = m \Leftrightarrow 2x_{0}^{2}-4x_{0}+4-m$ (1)

Because there is a unique value for $x_{0}$ so the equation (1) must have only one solution:

$\Leftrightarrow \Delta_{(1)} = 0 \Leftrightarrow 16-4 \times 2(4-m)=0 \Leftrightarrow m=2$ .

Plug $m=2$ into the equation system, we have:

$\begin{cases} (x-2)^{2}+y^{2}=2 \\ (y-2)^{2}+x^{2}=2 \end{cases}$

$\implies (x-2)^{2}+y^{2}+(y-2)^{2}+x^{2}=4 \Leftrightarrow (x-1)^{2}+(y-1)^{2}=0 \Leftrightarrow x=y=1$

We see that the solution satisfies the condition, so $\boxed{m=2}$ .

Both the given equations represent circles; one with centre $(0,2)$ , other with centre $(2,0)$ .

According to the condition—that the system should have only one solution—both circle have to be tangent to each other. Also, both circles have same radius.

For two such circles, the line $y=x$ is also tangent to both the circles at the same point of contact. Now, $m$ being the square of the radius of either circle, and the radius being the perpendicular distance of the centre (say $(0,2)$ ) from the line $y=x$ , we find $m=2$ .