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Difficulty: HardEquations of Circles in the Coordinate Plane

The equation x2+y2+10x−6y−15=0x^2 + y^2 + 10x - 6y - 15 = 0 represents a circle in the coordinate plane. If this circle is translated 22 units to the right and 44 units down, which of the following equations represents the translated circle?

  1. A
    (x−3)2+(y−1)2=49(x - 3)^2 + (y - 1)^2 = 49
  2. (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49Answer
  3. C
    (x+3)2+(y+1)2=7(x + 3)^2 + (y + 1)^2 = 7
  4. D
    (x+7)2+(y−7)2=49(x + 7)^2 + (y - 7)^2 = 49

Answer

The equation of the translated circle is (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.
The correct answer shows (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49. Completing the square on the original equation reveals the original center is (−5,3)(-5, 3) and r2=49r^2 = 49. Translating the center 22 units right and 44 units down shifts the xx-coordinate from −5-5 to −3-3 and the yy-coordinate from 33 to −1-1. The equation for a circle with center (−3,−1)(-3, -1) and radius squared of 4949 is (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.

Step-by-Step Solution

1
Group the xx-terms and yy-terms together, and move the constant term to the right side of the equation.
(x2+10x)+(y2−6y)=15(x^2 + 10x) + (y^2 - 6y) = 15
This groups terms by variable to prepare for completing the square.
2
Complete the square for the xx-terms by adding (102)2=25(\frac{10}{2})^2 = 25 and for the yy-terms by adding (−62)2=9(\frac{-6}{2})^2 = 9 to both sides of the equation.
(x2+10x+25)+(y2−6y+9)=15+25+9(x^2 + 10x + 25) + (y^2 - 6y + 9) = 15 + 25 + 9, which simplifies to (x+5)2+(y−3)2=49(x + 5)^2 + (y - 3)^2 = 49.
This converts the circle's equation into standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Identify the center and the radius squared of the original circle from the standard form equation.
The center is (h,k)=(−5,3)(h, k) = (-5, 3) and the radius squared is r2=49r^2 = 49.
The standard form reveals the center coordinates as the opposite signs of the constants inside the parentheses, and the right side is r2r^2.
4
Calculate the center of the translated circle by shifting the original center (−5,3)(-5, 3) by 22 units to the right and 44 units down.
The new center coordinates are (−5+2,3−4)=(−3,−1)(-5 + 2, 3 - 4) = (-3, -1).
Translating a point right increases its xx-coordinate, and translating it down decreases its yy-coordinate.
5
Write the standard form equation for the new circle with center (−3,−1)(-3, -1) and the unchanged radius squared r2=49r^2 = 49.
(x−(−3))2+(y−(−1))2=49(x - (-3))^2 + (y - (-1))^2 = 49, which simplifies to (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.
Substituting h=−3h = -3 and k=−1k = -1 into the standard circle equation form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 yields the final equation.

Key Concept

Completing the square to find standard circle equations and translating circle centers in the coordinate plane.
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Topics

Equations of Circles in the Coordinate Plane

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