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Difficulty: MediumSystems of Linear and Non-Linear Equations

A circle in the standard (x,y)(x, y) coordinate plane is defined by the equation (x−1)2+(y−2)2=10(x - 1)^2 + (y - 2)^2 = 10, and a line is defined by the equation y=3x−1y = 3x - 1. The line intersects the circle at two points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). What is the sum of the yy-coordinates of these two points of intersection?

Answer: 4

Answer

The sum of the yy-coordinates of the intersection points is 4.
Substituting y=3x−1y = 3x - 1 into the circle's equation gives (x−1)2+(3x−3)2=10(x-1)^2 + (3x-3)^2 = 10. Factoring out 3 from the second term yields (x−1)2+9(x−1)2=10(x-1)^2 + 9(x-1)^2 = 10, which simplifies to 10(x−1)2=1010(x-1)^2 = 10, or (x−1)2=1(x-1)^2 = 1. Solving for xx gives x=0x = 0 and x=2x = 2. Substituting these values into the linear equation gives the corresponding yy-coordinates: y=−1y = -1 when x=0x = 0, and y=5y = 5 when x=2x = 2. The sum of these yy-coordinates is 5+(−1)=45 + (-1) = 4.

Step-by-Step Solution

1
Substitute the expression for yy from the linear equation into the circle's equation.
(x−1)2+(3x−3)2=10(x-1)^2 + (3x-3)^2 = 10
To reduce the system of two equations to a single equation in terms of xx.
2
Factor out 3 from the term (3x−3)(3x-3) and simplify the equation.
10(x−1)2=1010(x-1)^2 = 10, which simplifies to (x−1)2=1(x-1)^2 = 1
To solve for the xx-coordinates of the intersection points.
3
Solve the simplified quadratic equation for xx.
x1=2x_1 = 2 and x2=0x_2 = 0
To find the xx-coordinates of the two intersection points.
4
Substitute the xx-values back into the linear equation y=3x−1y = 3x - 1 to find the corresponding yy-coordinates.
y1=3(2)−1=5y_1 = 3(2) - 1 = 5 and y2=3(0)−1=−1y_2 = 3(0) - 1 = -1
To determine the yy-coordinates of the intersection points (2,5)(2, 5) and (0,−1)(0, -1).
5
Add the two yy-coordinates together.
5+(−1)=45 + (-1) = 4
To find the sum of the yy-coordinates as requested by the question.

Key Concept

Solving systems of linear and circular equations by substitution

Alternative Method

Instead of finding the individual coordinates, substitute y1=3x1−1y_1 = 3x_1 - 1 and y2=3x2−1y_2 = 3x_2 - 1 to write the sum as y1+y2=3(x1+x2)−2y_1 + y_2 = 3(x_1 + x_2) - 2. Expanding the substitution equation gives 10x2−20x=010x^2 - 20x = 0. By Vieta's formulas, the sum of the roots x1+x2=−(−20)/10=2x_1 + x_2 = -(-20)/10 = 2. Substituting this back gives y1+y2=3(2)−2=4y_1 + y_2 = 3(2) - 2 = 4.
Estimated Time:1m 30s
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Topics

Systems of Linear and Non-Linear Equations

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