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Difficulty: MediumSystems of Linear and Non-Linear Equations

A circle in the standard (x,y)(x, y) coordinate plane is described by the equation x2+y2=68x^2 + y^2 = 68. A line is described by the equation y=x−6y = x - 6. The line intersects the circle at two points. What is the distance between these two points of intersection?

  1. A
    828\sqrt{2}
  2. B
    2172\sqrt{17}
  3. 10210\sqrt{2}Answer
  4. D
    2020
  5. E
    626\sqrt{2}

Answer

The distance between the two points of intersection is 10210\sqrt{2}.
To find the points of intersection, substitute y=x−6y = x - 6 into x2+y2=68x^2 + y^2 = 68, which yields x2+(x−6)2=68x^2 + (x - 6)^2 = 68. Expanding the binomial correctly results in x2+x2−12x+36=68x^2 + x^2 - 12x + 36 = 68. Combining like terms and writing the equation in standard form gives 2x2−12x−32=02x^2 - 12x - 32 = 0. Dividing the entire equation by 2 gives x2−6x−16=0x^2 - 6x - 16 = 0. Factoring this quadratic equation yields (x−8)(x+2)=0(x - 8)(x + 2) = 0, giving solutions x=8x = 8 and x=−2x = -2. Substituting these values back into the linear equation gives the points (8,2)(8, 2) and (−2,−8)(-2, -8). Finally, using the distance formula, the distance between the two points is (8−(−2))2+(2−(−8))2=102+102=200=102\sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}.

Step-by-Step Solution

1
Substitute the linear equation into the circle equation.
x2+(x−6)2=68x^2 + (x - 6)^2 = 68
To eliminate yy and solve for the xx-coordinates of the intersection points.
2
Expand the binomial and collect like terms.
2x2−12x−32=02x^2 - 12x - 32 = 0
Expanding (x−6)2(x - 6)^2 yields x2−12x+36x^2 - 12x + 36, and combining it with x2x^2 and subtracting 6868 from both sides puts the equation in quadratic form.
3
Divide the quadratic equation by 2 and solve by factoring.
(x−8)(x+2)=0(x - 8)(x + 2) = 0, so x=8x = 8 or x=−2x = -2
Simplifying the equation to x2−6x−16=0x^2 - 6x - 16 = 0 makes it easy to find the roots of the quadratic equation.
4
Find the corresponding yy-coordinates by substituting the xx-values back into the linear equation.
For x=8x = 8, y=2y = 2, yielding point (8,2)(8, 2). For x=−2x = -2, y=−8y = -8, yielding point (−2,−8)(-2, -8).
To determine the full coordinates of the two intersection points.
5
Apply the distance formula to find the distance between (8,2)(8, 2) and (−2,−8)(-2, -8).
d=(8−(−2))2+(2−(−8))2=102+102=200=102d = \sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}
To calculate the straight-line distance between the two coordinate points.

Key Concept

Solving systems of linear and non-linear equations by substitution and finding the distance between intersection points.
Estimated Time:1m 30s
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Topics

Systems of Linear and Non-Linear Equations

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