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Difficulty: MediumNonlinear Systems of Equations

A circle and a line intersect at two points in the xyxy-plane. The equations of the circle and the line are:

(x−3)2+y2=25(x - 3)^2 + y^2 = 25
y=x−2y = x - 2

If (x1,y1)(x_1, y_1) is the intersection point in Quadrant I and (x2,y2)(x_2, y_2) is the intersection point in Quadrant III, what is the value of x1+y2x_1 + y_2?

  1. 3Answer
  2. B
    5
  3. C
    1
  4. D
    10

Answer

The value of x1+y2x_1 + y_2 is 3.
To solve the system, we substitute the linear equation y=x−2y = x - 2 into the circle equation, obtaining (x−3)2+(x−2)2=25(x - 3)^2 + (x - 2)^2 = 25. Expanding this gives x2−6x+9+x2−4x+4=25x^2 - 6x + 9 + x^2 - 4x + 4 = 25, which simplifies to 2x2−10x−12=02x^2 - 10x - 12 = 0. Dividing by 2, we get x2−5x−6=0x^2 - 5x - 6 = 0, which factors as (x−6)(x+1)=0(x - 6)(x + 1) = 0. The solutions are x=6x = 6 and x=−1x = -1. Substituting these back into the linear equation gives the points of intersection: (6,4)(6, 4) in Quadrant I (since both coordinates are positive) and (−1,−3)(-1, -3) in Quadrant III (since both coordinates are negative). Therefore, x1=6x_1 = 6 and y2=−3y_2 = -3, and their sum is 6+(−3)=36 + (-3) = 3. This matches the correct option.

Step-by-Step Solution

1
Substitute the linear equation into the circle equation to obtain an equation in terms of xx only.
(x−3)2+(x−2)2=25(x - 3)^2 + (x - 2)^2 = 25
This reduces the system to a single quadratic equation in one variable, allowing us to find the x-coordinates of the intersection points.
2
Expand and simplify the quadratic equation into standard form.
x2−6x+9+x2−4x+4=25  ⟹  2x2−10x−12=0  ⟹  x2−5x−6=0x^2 - 6x + 9 + x^2 - 4x + 4 = 25 \implies 2x^2 - 10x - 12 = 0 \implies x^2 - 5x - 6 = 0
Simplifying the equation makes it solvable by standard quadratic factoring methods.
3
Solve the simplified quadratic equation for xx by factoring.
(x−6)(x+1)=0  ⟹  x=6(x - 6)(x + 1) = 0 \implies x = 6 or x=−1x = -1
The roots of this quadratic equation are the x-coordinates of the two intersection points.
4
Find the corresponding yy-coordinates by substituting the xx-values back into the linear equation y=x−2y = x - 2.
For x=6x = 6, y=6−2=4y = 6 - 2 = 4, giving the point (6,4)(6, 4) in Quadrant I. For x=−1x = -1, y=−1−2=−3y = -1 - 2 = -3, giving the point (−1,−3)(-1, -3) in Quadrant III.
This yields the coordinates of the two points of intersection and identifies which point lies in which quadrant.
5
Identify x1x_1 and y2y_2 and calculate the sum x1+y2x_1 + y_2.
x1=6x_1 = 6 and y2=−3y_2 = -3, so x1+y2=6+(−3)=3x_1 + y_2 = 6 + (-3) = 3.
This answers the specific question using the coordinate components of the two points.

Key Concept

Nonlinear Systems of Equations
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Topics

Nonlinear Systems of Equations

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