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Difficulty: HardNonlinear Systems of Equations
(x−4)2+(y+1)2=18x+y=k\begin{aligned} (x - 4)^2 + (y + 1)^2 &= 18 \\ x + y &= k \end{aligned}

In the system of equations above, kk is a constant. If the system has exactly one real solution, what is the sum of all possible values of kk?

  1. A
    -6
  2. B
    2
  3. 6Answer
  4. D
    10

Answer

The sum of all possible values of kk is 66.
The system of equations has exactly one real solution when the line is tangent to the circle. The circle (x−4)2+(y+1)2=18(x - 4)^2 + (y + 1)^2 = 18 has center (4,−1)(4, -1) and radius 18=32\sqrt{18} = 3\sqrt{2}. The distance from the center of the circle to the line x+y−k=0x + y - k = 0 is given by ∣4−1−k∣12+12=∣3−k∣2\frac{|4 - 1 - k|}{\sqrt{1^2 + 1^2}} = \frac{|3 - k|}{\sqrt{2}}. Setting this distance equal to the radius yields ∣3−k∣2=32  ⟹  ∣3−k∣=6\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6. Solving this equation gives k=9k = 9 and k=−3k = -3. The sum of these possible values is 9+(−3)=69 + (-3) = 6.

Step-by-Step Solution

1
Relate the algebraic condition of having exactly one solution to the geometric relationship between the circle and the line.
The line must be tangent to the circle, meaning the perpendicular distance from the center of the circle to the line must equal the radius of the circle.
A system consisting of a circle and a line has exactly one real solution if and only if the line is tangent to the circle.
2
Identify the center and radius of the circle from the equation (x−4)2+(y+1)2=18(x - 4)^2 + (y + 1)^2 = 18.
The center of the circle is (4,−1)(4, -1) and the radius is r=18=32r = \sqrt{18} = 3\sqrt{2}.
The standard form of a circle equation is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.
3
Express the line equation in standard form Ax+By+C=0Ax + By + C = 0 and set up the perpendicular distance formula.
The line equation is x+y−k=0x + y - k = 0. The distance from the center (4,−1)(4, -1) to the line is d=∣(1)(4)+(1)(−1)−k∣12+12=∣3−k∣2d = \frac{|(1)(4) + (1)(-1) - k|}{\sqrt{1^2 + 1^2}} = \frac{|3 - k|}{\sqrt{2}}.
The perpendicular distance from a point (x0,y0)(x_0, y_0) to a line Ax+By+C=0Ax + By + C = 0 is given by d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.
4
Set the distance equal to the radius of the circle and solve the absolute value equation for kk.
\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6 .Thisyieldstwoequations:. This yields two equations: 3 - k = 6 \implies k = -3 ,and, and 3 - k = -6 \implies k = 9$.
Setting the distance equal to the radius ensures that the line is tangent to the circle, resulting in exactly one point of intersection.
5
Calculate the sum of all possible values of kk.
Sum = −3+9=6-3 + 9 = 6.
The question asks for the sum of all values of kk that satisfy the tangency condition.

Key Concept

Determining the relationship between a line and a circle in a system of equations by setting the perpendicular distance from the center of the circle equal to its radius.

Alternative Method

Substitute y=k−xy = k - x into the circle equation to obtain a quadratic equation in xx: (x−4)2+(k−x+1)2=18(x - 4)^2 + (k - x + 1)^2 = 18. Expanding and combining like terms yields 2x2−2(k+5)x+(k2+2k−1)=02x^2 - 2(k + 5)x + (k^2 + 2k - 1) = 0. For the system to have exactly one real solution, the discriminant Δ\Delta of this quadratic equation must be zero: Δ=[−2(k+5)]2−4(2)(k2+2k−1)=0\Delta = [-2(k + 5)]^2 - 4(2)(k^2 + 2k - 1) = 0. Simplifying this equation leads to 4(k2+10k+25)−8(k2+2k−1)=0  ⟹  k2−6k−27=04(k^2 + 10k + 25) - 8(k^2 + 2k - 1) = 0 \implies k^2 - 6k - 27 = 0. By Vieta's formulas, the sum of the solutions to this quadratic equation for kk is given by −ba=−−61=6-\frac{b}{a} = -\frac{-6}{1} = 6.
Estimated Time:2m 30s
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Topics

Nonlinear Systems of Equations

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