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Zorluk: OrtaSystems of Linear and Non-Linear Equations

A circular search-and-rescue radar zone centered at a local station is modeled by the equation x2+y2=25x^2 + y^2 = 25 in the standard (x,y)(x, y) coordinate plane, where coordinates are measured in miles. A rescue helicopter flies along a straight path modeled by the line y=2x−5y = 2x - 5. What is the distance, in miles, the helicopter travels through the radar zone?

  1. A
    252\sqrt{5}
  2. 454\sqrt{5}Cevap
  3. C
    88
  4. D
    1010
  5. E
    10210\sqrt{2}

Cevap

The distance the helicopter travels through the radar zone is 454\sqrt{5} miles.
To find the distance the helicopter travels through the radar zone, we must determine the distance between the two points of intersection of the circular boundary x2+y2=25x^2 + y^2 = 25 and the line y=2x−5y = 2x - 5. Substituting the expression for yy into the circular equation yields x2+(2x−5)2=25x^2 + (2x-5)^2 = 25. Expanding the binomial correctly gives x2+4x2−20x+25=25x^2 + 4x^2 - 20x + 25 = 25, which simplifies to 5x2−20x=05x^2 - 20x = 0. Factoring this equation as 5x(x−4)=05x(x-4) = 0 gives x=0x = 0 and x=4x = 4. Substituting these values back into the linear equation yields the points of intersection (0,−5)(0, -5) and (4,3)(4, 3). The distance between these two points is (4−0)2+(3−(−5))2=16+64=80=45\sqrt{(4-0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} miles.

Adım Adım Çözüm

1
Substitute the linear equation y=2x−5y = 2x - 5 into the circular equation x2+y2=25x^2 + y^2 = 25.
x2+(2x−5)2=25x^2 + (2x - 5)^2 = 25
To find the coordinates of the intersection points where the helicopter's path meets the boundary of the radar zone.
2
Expand the binomial (2x−5)2(2x - 5)^2 and simplify the quadratic equation.
x2+4x2−20x+25=25  ⟹  5x2−20x=0x^2 + 4x^2 - 20x + 25 = 25 \implies 5x^2 - 20x = 0
To collect like terms and put the equation in a solvable quadratic form.
3
Factor the quadratic equation 5x2−20x=05x^2 - 20x = 0 to solve for xx.
5x(x−4)=0  ⟹  x=0 or x=45x(x - 4) = 0 \implies x = 0 \text{ or } x = 4
To determine the xx-coordinates of the two intersection points.
4
Determine the corresponding yy-coordinates by substituting the xx-values into the linear equation y=2x−5y = 2x - 5.
For x=0x = 0: y=2(0)−5=−5  ⟹  (0,−5)y = 2(0) - 5 = -5 \implies (0, -5). For x=4x = 4: y=2(4)−5=3  ⟹  (4,3)y = 2(4) - 5 = 3 \implies (4, 3).
To obtain the exact coordinate pairs for the entry and exit points.
5
Use the distance formula to calculate the distance between the two points (0,−5)(0, -5) and (4,3)(4, 3).
d=(4−0)2+(3−(−5))2=16+64=80=45d = \sqrt{(4 - 0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}
To compute the straight-line distance traveled by the helicopter through the radar zone.

Anahtar Kavram

Solving systems of linear and circular equations by substitution and finding the distance between their intersection points.
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Konular

Systems of Linear and Non-Linear Equations

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