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Zorluk: OrtaSystems of Linear and Non-Linear Equations

In the standard (x,y)(x, y) coordinate plane, the circle defined by x2+y2=26x^2 + y^2 = 26 and the line defined by y=x−4y = x - 4 intersect at two points. What is the distance between these two points of intersection?

  1. A
    2102\sqrt{10}
  2. B
    424\sqrt{2}
  3. C
    66
  4. 626\sqrt{2}Cevap
  5. E
    1212

Cevap

626\sqrt{2}
Solving the system of equations by substituting y=x−4y = x - 4 into x2+y2=26x^2 + y^2 = 26 yields 2x2−8x−10=02x^2 - 8x - 10 = 0, which simplifies to x2−4x−5=0x^2 - 4x - 5 = 0. Factoring gives (x−5)(x+1)=0(x-5)(x+1) = 0, leading to the points (5,1)(5, 1) and (−1,−5)(-1, -5). The distance between these points is 62+62=62\sqrt{6^2 + 6^2} = 6\sqrt{2}.

Adım Adım Çözüm

1
Substitute the linear equation into the circle equation.
x2+(x−4)2=26x^2 + (x - 4)^2 = 26
This reduces the system to a single equation in terms of xx.
2
Expand the squared binomial and simplify the quadratic equation.
2x2−8x−10=02x^2 - 8x - 10 = 0, which simplifies to x2−4x−5=0x^2 - 4x - 5 = 0.
Expanding (x−4)2(x - 4)^2 yields x2−8x+16x^2 - 8x + 16. Setting the equation to zero allows us to solve for xx.
3
Factor the quadratic equation to find the xx-coordinates.
x=5x = 5 or x=−1x = -1.
The factored form is (x−5)(x+1)=0(x - 5)(x + 1) = 0.
4
Substitute the xx-values back into the linear equation to find the corresponding yy-coordinates.
The intersection points are (5,1)(5, 1) and (−1,−5)(-1, -5).
For x=5x = 5, y=5−4=1y = 5 - 4 = 1. For x=−1x = -1, y=−1−4=−5y = -1 - 4 = -5.
5
Apply the distance formula to find the straight-line distance between the two points.
626\sqrt{2}
The distance is (5−(−1))2+(1−(−5))2=62+62=72=62\sqrt{(5 - (-1))^2 + (1 - (-5))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}.

Anahtar Kavram

Solving systems of linear and non-linear (circular) equations by substitution and finding the distance between intersection points.
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Konular

Systems of Linear and Non-Linear Equations

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