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Zorluk: OrtaEquations and Graphs of Circles

A circle in the standard (x,y)(x, y) coordinate plane is represented by the equation x2+y2+10x−6y+9=0x^2 + y^2 + 10x - 6y + 9 = 0. If this circle is translated 44 units to the right and 22 units down, which of the following is the equation of the translated circle?

  1. (x+1)2+(y−1)2=25(x + 1)^2 + (y - 1)^2 = 25Cevap
  2. B
    (x−1)2+(y+1)2=25(x - 1)^2 + (y + 1)^2 = 25
  3. C
    (x+1)2+(y−1)2=5(x + 1)^2 + (y - 1)^2 = 5
  4. D
    (x+9)2+(y−5)2=25(x + 9)^2 + (y - 5)^2 = 25
  5. E
    (x+1)2+(y−1)2=9(x + 1)^2 + (y - 1)^2 = 9

Cevap

(x+1)2+(y−1)2=25(x + 1)^2 + (y - 1)^2 = 25
Completing the square on x2+y2+10x−6y+9=0x^2 + y^2 + 10x - 6y + 9 = 0 yields (x+5)2+(y−3)2=25(x + 5)^2 + (y - 3)^2 = 25, identifying the original center as (−5,3)(-5, 3) and r2=25r^2 = 25. Shifting the center 44 units right and 22 units down moves (−5,3)(-5, 3) to (−5+4,3−2)=(−1,1)(-5 + 4, 3 - 2) = (-1, 1). Substituting the updated center (−1,1)(-1, 1) and r2=25r^2 = 25 into the standard equation of a circle (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 gives (x+1)2+(y−1)2=25(x + 1)^2 + (y - 1)^2 = 25.

Adım Adım Çözüm

1
Group terms and complete the square for both xx and yy variables in the original equation.
(x+5)2+(y−3)2=25(x + 5)^2 + (y - 3)^2 = 25
Rearrange x2+y2+10x−6y+9=0x^2 + y^2 + 10x - 6y + 9 = 0 as (x2+10x)+(y2−6y)=−9(x^2 + 10x) + (y^2 - 6y) = -9. Add (10/2)2=25(10/2)^2 = 25 and (−6/2)2=9(-6/2)^2 = 9 to both sides: (x+5)2+(y−3)2=−9+25+9=25(x + 5)^2 + (y - 3)^2 = -9 + 25 + 9 = 25.
2
Determine the center and radius of the original circle.
Original center is (−5,3)(-5, 3) and r2=25r^2 = 25 (radius r=5r = 5).
Comparing (x+5)2+(y−3)2=25(x + 5)^2 + (y - 3)^2 = 25 to (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 gives h=−5h = -5, k=3k = 3, and r2=25r^2 = 25.
3
Calculate the coordinates of the new center after the horizontal and vertical translation.
New center is (−1,1)(-1, 1).
Translating 44 units to the right adds 44 to the xx-coordinate: −5+4=−1-5 + 4 = -1. Translating 22 units down subtracts 22 from the yy-coordinate: 3−2=13 - 2 = 1.
4
Substitute the new center (−1,1)(-1, 1) and unchanged radius squared value r2=25r^2 = 25 into standard form.
(x+1)2+(y−1)2=25(x + 1)^2 + (y - 1)^2 = 25
Using (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 with h=−1h = -1 and k=1k = 1 gives (x−(−1))2+(y−1)2=25(x - (-1))^2 + (y - 1)^2 = 25, which simplifies to (x+1)2+(y−1)2=25(x + 1)^2 + (y - 1)^2 = 25.

Anahtar Kavram

Converting circle equations to standard form by completing the square and performing translations in the coordinate plane.
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Konular

Equations and Graphs of Circles

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