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Difficulty: MediumEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, a circle is defined by the equation x2+y2+14x−6y+9=0x^2 + y^2 + 14x - 6y + 9 = 0. What is the area, in square units, of the region enclosed by the circle?

  1. A
    7π7\pi
  2. B
    14π14\pi
  3. C
    40π40\pi
  4. 49π49\piAnswer
  5. E
    58π58\pi

Answer

The area of the region enclosed by the circle is 49π49\pi square units.
To find the area of the circle, we rewrite the equation x2+y2+14x−6y+9=0x^2 + y^2 + 14x - 6y + 9 = 0 in standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. Regrouping gives (x2+14x)+(y2−6y)=−9(x^2 + 14x) + (y^2 - 6y) = -9. Completing the square for xx requires adding (142)2=49(\frac{14}{2})^2 = 49, and for yy requires adding (−62)2=9(\frac{-6}{2})^2 = 9. Adding both to the right side gives −9+49+9=49-9 + 49 + 9 = 49. Thus, the standard equation is (x+7)2+(y−3)2=49(x + 7)^2 + (y - 3)^2 = 49, yielding r2=49r^2 = 49. The area enclosed by the circle is πr2=49π\pi r^2 = 49\pi.

Step-by-Step Solution

1
Group terms and move the constant term to the right side of the equation.
(x2+14x)+(y2−6y)=−9(x^2 + 14x) + (y^2 - 6y) = -9
Grouping xx-terms and yy-terms prepares the equation for completing the square.
2
Complete the square for both xx and yy expressions by adding (142)2=49(\frac{14}{2})^2 = 49 and (−62)2=9(\frac{-6}{2})^2 = 9 to both sides.
(x2+14x+49)+(y2−6y+9)=−9+49+9(x^2 + 14x + 49) + (y^2 - 6y + 9) = -9 + 49 + 9
Adding these values turns the quadratic expressions into perfect square trinomials while keeping the equation balanced.
3
Write in standard circle equation form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 to identify r2r^2.
(x+7)2+(y−3)2=49(x + 7)^2 + (y - 3)^2 = 49, so r2=49r^2 = 49
The standard form directly gives the radius squared, r2=49r^2 = 49.
4
Calculate the area of the circle using the formula A=πr2A = \pi r^2.
A=π(49)=49πA = \pi (49) = 49\pi
Substituting r2=49r^2 = 49 into the area formula gives the final enclosed area.

Key Concept

Converting general circle equation x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 to standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 by completing the square.
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Topics

Equations and Graphs of Circles

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