Given that these simultaneous equations \( x-y=k \) \( x^{2}+y^{2}-9=0 \) have exactly one pair of solutions, show that \( k= \pm 3 \sqrt{2} \)

To determine the values of \( k \) for which the system of equations \[ \begin{cases} x - y = k \\ x^2 + y^2 = 9 \end{cases} \] has exactly one pair of solutions, we can approach the problem both geometrically and algebraically. ### **Geometric Interpretation:** 1. **Equation of the Line:** \[ x - y = k \] This represents a straight line with slope \( 1 \). 2. **Equation of the Circle:** \[ x^2 + y^2 = 9 \] This is a circle centered at the origin \((0, 0)\) with radius \( 3 \). For the system to have exactly one solution, the line must be **tangent** to the circle. The condition for tangency is that the perpendicular distance from the center of the circle to the line equals the radius of the circle. 3. **Calculating the Distance:** The distance \( d \) from the center \((0, 0)\) to the line \( x - y = k \) is given by: \[ d = \frac{|0 - 0 - k|}{\sqrt{1^2 + (-1)^2}} = \frac{|k|}{\sqrt{2}} \] Setting this equal to the radius: \[ \frac{|k|}{\sqrt{2}} = 3 \implies |k| = 3\sqrt{2} \] Therefore, \( k = \pm 3\sqrt{2} \). ### **Algebraic Approach:** 1. **Express \( x \) in terms of \( y \):** \[ x = y + k \] 2. **Substitute into the Circle's Equation:** \[ (y + k)^2 + y^2 = 9 \] Expanding: \[ y^2 + 2ky + k^2 + y^2 = 9 \implies 2y^2 + 2ky + (k^2 - 9) = 0 \] 3. **Condition for One Solution:** For the quadratic equation in \( y \) to have exactly one solution, the discriminant must be zero: \[ D = (2k)^2 - 4 \cdot 2 \cdot (k^2 - 9) = 4k^2 - 8k^2 + 72 = -4k^2 + 72 = 0 \] Solving: \[ -4k^2 + 72 = 0 \implies 4k^2 = 72 \implies k^2 = 18 \implies k = \pm 3\sqrt{2} \] Both methods conclusively show that the system has exactly one pair of solutions only when \( k = \pm 3\sqrt{2} \). **Answer:** After simplifying, the only possible values are ± 3 times √2. Thus, k = ± 3 √2.