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Difficulty: HardEquations of Circles in the Coordinate Plane

In the xyxy-plane, the graph of the equation x2+y2−12x+8y−k=0x^2 + y^2 - 12x + 8y - k = 0, where kk is a positive constant, is a circle. If the line y=6y = 6 is tangent to the circle, what is the value of kk?

Answer: 48

Answer

48
To find the value of kk, we convert the given circle equation into its standard form, (x−h)2+(y−j)2=r2(x - h)^2 + (y - j)^2 = r^2. Grouping the terms gives (x2−12x)+(y2+8y)=k(x^2 - 12x) + (y^2 + 8y) = k. Completing the square for xx and yy gives (x−6)2−36+(y+4)2−16=k(x - 6)^2 - 36 + (y + 4)^2 - 16 = k, which simplifies to (x−6)2+(y+4)2=k+52(x - 6)^2 + (y + 4)^2 = k + 52. This tells us that the center of the circle is (6,−4)(6, -4) and the radius squared is r2=k+52r^2 = k + 52. A horizontal line y=6y = 6 is tangent to the circle, meaning the perpendicular distance from the center (6,−4)(6, -4) to the line y=6y = 6 is equal to the radius. This distance is ∣6−(−4)∣=10|6 - (-4)| = 10. Therefore, the radius is 1010, which means r2=100r^2 = 100. Equating the two expressions for the radius squared gives k+52=100k + 52 = 100. Solving this equation yields k=48k = 48.

Step-by-Step Solution

1
Group the variables and complete the square for the xx and yy terms.
(x−6)2+(y+4)2=k+52(x - 6)^2 + (y + 4)^2 = k + 52
Completing the square converts the equation from general form to standard form, which reveals the center and radius.
2
Identify the center of the circle and the algebraic representation of the radius.
Center is (6,−4)(6, -4) and radius r=k+52r = \sqrt{k + 52}.
The standard equation of a circle is (x−h)2+(y−j)2=r2(x - h)^2 + (y - j)^2 = r^2, where (h,j)(h, j) is the center and rr is the radius.
3
Find the radius of the circle using the given tangent line.
Radius r=10r = 10
The distance from the center's yy-coordinate, −4-4, to the horizontal tangent line y=6y = 6 is ∣6−(−4)∣=10|6 - (-4)| = 10, which represents the radius of the circle.
4
Equate the radius squared value to the algebraic expression for the radius squared and solve for kk.
k=48k = 48
Since the radius is 1010, the radius squared is 100100. Setting k+52=100k + 52 = 100 and solving for kk yields 4848.

Key Concept

Converting a circle's equation from general to standard form by completing the square, and using the distance from the center to a tangent line to find the radius.
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Topics

Equations of Circles in the Coordinate Plane

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