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Difficulty: HardEquations and Graphs of Circles

The circle defined by the equation x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0 is plotted in the standard (x,y)(x, y) coordinate plane. A second circle has a center that is the reflection of the first circle's center across the line y=xy = x. If the radius of the second circle is twice the radius of the first circle, which of the following is the equation of the second circle?

  1. A
    (x−4)2+(y+3)2=144(x - 4)^2 + (y + 3)^2 = 144
  2. B
    (x+4)2+(y−3)2=12(x + 4)^2 + (y - 3)^2 = 12
  3. C
    (x−3)2+(y+4)2=144(x - 3)^2 + (y + 4)^2 = 144
  4. (x+4)2+(y−3)2=144(x + 4)^2 + (y - 3)^2 = 144Answer
  5. E
    (x+4)2+(y−3)2=72(x + 4)^2 + (y - 3)^2 = 72

Answer

(x+4)2+(y−3)2=144(x + 4)^2 + (y - 3)^2 = 144
Completing the square on x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0 gives (x−3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36. The center is (3,−4)(3, -4) and the radius is 66. Reflecting (3,−4)(3, -4) across the line y=xy = x swaps the coordinates, yielding the new center (−4,3)(-4, 3). Doubling the radius yields 1212. The equation of the second circle is therefore (x−(−4))2+(y−3)2=122(x - (-4))^2 + (y - 3)^2 = 12^2, which simplifies to (x+4)2+(y−3)2=144(x + 4)^2 + (y - 3)^2 = 144.

Step-by-Step Solution

1
Group the xx and yy terms of the given equation x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0 and add 11 to both sides.
(x2−6x)+(y2+8y)=11(x^2 - 6x) + (y^2 + 8y) = 11
To prepare the terms for completing the square.
2
Complete the square for both the xx and yy groups by adding (6/2)2=9(6/2)^2 = 9 and (8/2)2=16(8/2)^2 = 16 to both sides.
(x−3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36
To express the circle's equation in standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Identify the center and radius of the first circle from the standard form.
Center is (3,−4)(3, -4) and radius is r1=36=6r_1 = \sqrt{36} = 6.
These characteristics are needed to find the properties of the second circle.
4
Reflect the center (3,−4)(3, -4) across the line y=xy = x by swapping the xx and yy coordinates, and double the radius.
New center is (−4,3)(-4, 3) and new radius is r2=2×6=12r_2 = 2 \times 6 = 12.
To satisfy the geometric transformations specified in the problem statement.
5
Write the standard equation of the second circle using the new center (−4,3)(-4, 3) and new radius 1212.
(x+4)2+(y−3)2=144(x + 4)^2 + (y - 3)^2 = 144
Substituting h=−4h = -4, k=3k = 3, and r=12r = 12 into the standard equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

Key Concept

Completing the square to find standard circle characteristics, and applying reflections across the line y=xy = x.
Estimated Time:2m 0s
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Topics

Equations and Graphs of Circles

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