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Difficulty: Very hardEquations and Graphs of Circles

An equation of a circle is given by x2+y2−8x+6y+k=0x^2 + y^2 - 8x + 6y + k = 0, where kk is a constant. A second circle has a center that is the reflection of the first circle's center across the line y=xy = x. If the second circle is tangent to the xx-axis and has the same radius as the first circle, what is the value of kk?

  1. 9Answer
  2. B
    16
  3. C
    21
  4. D
    34
  5. E
    41

Answer

The value of the constant kk is 9.
The correct answer is 9. Completing the square for the first circle's equation gives (x−4)2+(y+3)2=25−k(x - 4)^2 + (y + 3)^2 = 25 - k, which identifies the center as (4,−3)(4, -3) and the radius squared as r2=25−kr^2 = 25 - k. Reflecting (4,−3)(4, -3) across the line y=xy = x swaps the coordinates to give the new center (−3,4)(-3, 4). Because the second circle is tangent to the xx-axis, its radius is the absolute value of the yy-coordinate of its center, which is ∣4∣=4|4| = 4. Since both circles have the same radius, we set the radius squared equal to 424^2: 25−k=1625 - k = 16, which yields k=9k = 9.

Step-by-Step Solution

1
Complete the square for the first circle's equation x2+y2−8x+6y+k=0x^2 + y^2 - 8x + 6y + k = 0 to identify its center and radius.
(x−4)2+(y+3)2=25−k(x - 4)^2 + (y + 3)^2 = 25 - k, which represents a circle with center (4,−3)(4, -3) and radius squared r2=25−kr^2 = 25 - k.
This puts the equation into the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 to find the center and radius.
2
Reflect the center of the first circle across the line y=xy = x.
Reflecting the point (4,−3)(4, -3) across the line y=xy = x swaps the coordinates, yielding the new center (−3,4)(-3, 4).
To find the center of the second circle.
3
Determine the radius of the second circle using the tangency condition.
Since the second circle is tangent to the xx-axis, its radius is equal to the absolute value of the yy-coordinate of its center, which is ∣4∣=4|4| = 4.
The distance from a circle's center (h,k)(h, k) to the line of tangency y=0y = 0 (the xx-axis) is equal to its radius.
4
Equate the radius squared of the first circle to the square of the radius of the second circle.
25−k=42  ⟹  25−k=16  ⟹  k=925 - k = 4^2 \implies 25 - k = 16 \implies k = 9.
Both circles are stated to have the same radius.

Key Concept

Converting the general form of a circle's equation to standard form by completing the square, and using coordinate transformations and geometric tangency conditions to solve for unknowns.
Estimated Time:3m 0s
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Topics

Equations and Graphs of Circles

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