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Difficulty: Very hardEquations and Graphs of Circles

A circle in the standard (x,y)(x,y) coordinate plane is defined by the equation 3x2+3y2−24x+12y+k=03x^2 + 3y^2 - 24x + 12y + k = 0, where kk is a constant. If the circle is tangent to the line y=2y = 2, what is the value of kk?

  1. A
    -48
  2. B
    0
  3. 12Answer
  4. D
    48
  5. E
    60

Answer

12
The correct answer is 12. Dividing the given equation 3x2+3y2−24x+12y+k=03x^2 + 3y^2 - 24x + 12y + k = 0 by 33 gives x2+y2−8x+4y+k3=0x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0. Completing the square for both variables yields (x−4)2+(y+2)2=20−k3(x-4)^2 + (y+2)^2 = 20 - \frac{k}{3}. This represents a circle centered at (4,−2)(4, -2) with a radius squared of r2=20−k3r^2 = 20 - \frac{k}{3}. A circle centered at (4,−2)(4, -2) that is tangent to the line y=2y = 2 has a radius equal to the vertical distance between y=−2y = -2 and y=2y = 2, which is ∣−2−2∣=4|-2 - 2| = 4. Thus, r2=16r^2 = 16. Setting 20−k3=1620 - \frac{k}{3} = 16 yields k3=4\frac{k}{3} = 4, which gives k=12k = 12.

Step-by-Step Solution

1
Divide the entire equation by 3 to normalize the coefficients of the quadratic terms.
x2+y2−8x+4y+k3=0x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0
The standard equation of a circle requires the coefficients of x2x^2 and y2y^2 to be 1.
2
Complete the square for the xx and yy terms.
(x−4)2−16+(y+2)2−4+k3=0⇒(x−4)2+(y+2)2=20−k3(x - 4)^2 - 16 + (y + 2)^2 - 4 + \frac{k}{3} = 0 \Rightarrow (x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3}
This rewrites the equation in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 to identify the center and radius.
3
Identify the center of the circle and set up the radius relationship.
Center is (4,−2)(4, -2) and the radius squared is r2=20−k3r^2 = 20 - \frac{k}{3}.
Comparing (x−4)2+(y+2)2=20−k3(x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3} to the standard form gives the center (h,k)=(4,−2)(h, k) = (4, -2) and r2r^2.
4
Determine the radius using the tangency condition with the line y=2y = 2.
Radius r=∣ycenter−2∣=∣−2−2∣=4r = |y_{\text{center}} - 2| = |-2 - 2| = 4
Since the circle is tangent to the horizontal line y=2y = 2, the radius is the vertical distance from the center's y-coordinate to the line.
5
Solve for the constant kk using the radius squared value.
r2=16⇒20−k3=16⇒4=k3⇒k=12r^2 = 16 \Rightarrow 20 - \frac{k}{3} = 16 \Rightarrow 4 = \frac{k}{3} \Rightarrow k = 12
Equating the two expressions for r2r^2 allows us to solve for kk.

Key Concept

Converting the general equation of a circle to standard form by completing the square and using geometric tangency conditions to solve for unknown constants.
Estimated Time:2m 0s
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Topics

Equations and Graphs of Circles

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