Question

Difficulty: MediumEquations of Circles in the Coordinate Plane

In the xyxy-plane, the graph of the equation x2+y212x+10y+k=0x^2 + y^2 - 12x + 10y + k = 0, where kk is a constant, is a circle. If this circle is tangent to the yy-axis, what is the value of kk?

Answer: 25

Answer

The value of kk is 25.
Completing the square for the equation x2+y212x+10y+k=0x^2 + y^2 - 12x + 10y + k = 0 yields standard form (x6)2+(y+5)2=61k(x - 6)^2 + (y + 5)^2 = 61 - k. The center of the circle is (6,5)(6, -5) and the radius squared is r2=61kr^2 = 61 - k. Since the circle is tangent to the yy-axis (the line x=0x = 0), its radius must be equal to the horizontal distance from the center to the yy-axis, which is the absolute value of the xx-coordinate of the center, 6=6|6| = 6. Therefore, the radius is 66, and the radius squared is r2=36r^2 = 36. Setting 61k=3661 - k = 36 gives k=25k = 25.

Step-by-Step Solution

1
Complete the square for xx and yy variables in the equation.
(x6)2+(y+5)2=61k(x - 6)^2 + (y + 5)^2 = 61 - k
Completing the square allows the general form equation of the circle to be written in standard form, where the center and radius are easily identifiable.
2
Determine the center and the expression for the radius squared from the standard form.
Center is (6,5)(6, -5) and r2=61kr^2 = 61 - k
In the standard circle equation (xh)2+(ykc)2=r2(x - h)^2 + (y - k_c)^2 = r^2, the center is (h,kc)(h, k_c) and the right side represents the radius squared.
3
Relate the tangency condition to the radius of the circle.
r=6r = 6
A circle tangent to the yy-axis has its radius equal to the distance from its center to the line x=0x = 0, which is the absolute value of the center's xx-coordinate.
4
Calculate the value of the constant kk.
k=25k = 25
Since r=6r = 6, then r2=36r^2 = 36. Setting 61k=3661 - k = 36 and solving for kk gives k=25k = 25.

Key Concept

Completing the square to find the standard form of a circle's equation and applying geometric tangency properties to determine coordinates or constants.
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