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Difficulty: MediumEquations and Graphs of Circles

A circle with center (5,−6)(5, -6) is tangent to the xx-axis in the standard (x,y)(x, y) coordinate plane. Which of the following is the equation of this circle?

  1. A
    x2+y2+10x−12y+25=0x^2 + y^2 + 10x - 12y + 25 = 0
  2. B
    x2+y2−10x+12y+36=0x^2 + y^2 - 10x + 12y + 36 = 0
  3. x2+y2−10x+12y+25=0x^2 + y^2 - 10x + 12y + 25 = 0Answer
  4. D
    x2+y2−25=0x^2 + y^2 - 25 = 0
  5. E
    x2+y2−10x+12y+55=0x^2 + y^2 - 10x + 12y + 55 = 0

Answer

x2+y2−10x+12y+25=0x^2 + y^2 - 10x + 12y + 25 = 0
The correct equation x2+y2−10x+12y+25=0x^2 + y^2 - 10x + 12y + 25 = 0 is obtained by identifying that a circle with center (5,−6)(5, -6) tangent to the xx-axis has a radius of ∣−6∣=6|-6| = 6. Substituting these into the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 yields (x−5)2+(y+6)2=36(x - 5)^2 + (y + 6)^2 = 36. Expanding the terms gives x2−10x+25+y2+12y+36=36x^2 - 10x + 25 + y^2 + 12y + 36 = 36, which simplifies to x2+y2−10x+12y+25=0x^2 + y^2 - 10x + 12y + 25 = 0.

Step-by-Step Solution

1
Determine the radius of the circle from the given center and tangency condition.
The center is (5,−6)(5, -6), and the circle is tangent to the xx-axis (the line y=0y = 0). The distance from the center to the xx-axis is the absolute value of the yy-coordinate of the center: ∣−6∣=6|-6| = 6. Therefore, the radius rr of the circle is 66.
A circle tangent to the xx-axis has a radius equal to the vertical distance from its center to the xx-axis.
2
Write the equation of the circle in standard form.
(x−5)2+(y−(−6))2=62(x - 5)^2 + (y - (-6))^2 = 6^2, which simplifies to (x−5)2+(y+6)2=36(x - 5)^2 + (y + 6)^2 = 36.
The standard equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.
3
Expand the standard equation to find the general quadratic form.
Expanding the squared binomials: (x2−10x+25)+(y2+12y+36)=36(x^2 - 10x + 25) + (y^2 + 12y + 36) = 36. Grouping terms and simplifying gives x2+y2−10x+12y+25=0x^2 + y^2 - 10x + 12y + 25 = 0.
To match the options, we convert the equation from standard form to general form by expanding the terms and setting the equation equal to zero.

Key Concept

The standard equation of a circle with center (h,k)(h, k) and radius rr is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. When a circle is tangent to the xx-axis, its radius is equal to the distance from the center to the xx-axis, which is ∣k∣|k|.
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Topics

Equations and Graphs of Circles

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