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Difficulty: HardEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, the circle defined by the equation x2+y2−12x−4y+31=0x^2 + y^2 - 12x - 4y + 31 = 0 is translated 33 units to the right and 55 units down. A line passing through the origin with a non-zero slope mm is tangent to this translated circle. What is the value of mm?

Answer: -0.75

Answer

The correct answer is -0.75.
The correct answer is -0.75. By completing the square on the original equation, we find the circle (x−6)2+(y−2)2=9(x-6)^2 + (y-2)^2 = 9 with center (6,2)(6, 2) and radius 33. Translating the circle shifts the center to (9,−3)(9, -3). A line passing through the origin with slope mm has the equation mx−y=0mx - y = 0. For this line to be tangent to the circle, its perpendicular distance from the center (9,−3)(9, -3) must equal the radius 33. Using the distance formula, we get the equation ∣9m+3∣m2+1=3\frac{|9m + 3|}{\sqrt{m^2 + 1}} = 3. Simplifying gives ∣3m+1∣=m2+1|3m + 1| = \sqrt{m^2 + 1}. Squaring both sides yields 9m2+6m+1=m2+19m^2 + 6m + 1 = m^2 + 1, which simplifies to 8m2+6m=08m^2 + 6m = 0. The non-zero solution is m=−0.75m = -0.75.

Step-by-Step Solution

1
Complete the square for the original circle equation x2+y2−12x−4y+31=0x^2 + y^2 - 12x - 4y + 31 = 0.
(x−6)2+(y−2)2=9(x - 6)^2 + (y - 2)^2 = 9.
Grouping the xx and yy terms and completing the square yields the standard form of a circle, showing that the original center is (6,2)(6, 2) and the radius r=9=3r = \sqrt{9} = 3.
2
Apply the translation of 33 units right and 55 units down to the center (6,2)(6, 2).
New center is (6+3,2−5)=(9,−3)(6 + 3, 2 - 5) = (9, -3).
Translating a point in the coordinate plane shifts the coordinates directly: xx-coordinate increases by 33 and yy-coordinate decreases by 55. The radius remains 33.
3
Set up the distance from the new center (9,−3)(9, -3) to the tangent line y=mxy = mx.
Distance equation: ∣9m−(−3)∣m2+1=3\frac{|9m - (-3)|}{\sqrt{m^2 + 1}} = 3.
The line y=mxy = mx can be rewritten as mx−y=0mx - y = 0. The perpendicular distance from the center to a tangent line must equal the radius of the circle, which is 33.
4
Simplify the distance equation and solve for mm.
∣3m+1∣=m2+1|3m + 1| = \sqrt{m^2 + 1}.
Factoring out 33 from the numerator and dividing both sides by 33 simplifies the equation before squaring.
5
Square both sides of the equation and solve the resulting quadratic equation.
2m(4m+3)=0  ⟹  m=02m(4m + 3) = 0 \implies m = 0 or m=−0.75m = -0.75.
Squaring both sides yields (3m+1)2=m2+1  ⟹  9m2+6m+1=m2+1  ⟹  8m2+6m=0(3m + 1)^2 = m^2 + 1 \implies 9m^2 + 6m + 1 = m^2 + 1 \implies 8m^2 + 6m = 0. Since we are looking for a non-zero slope, the only valid solution is m=−0.75m = -0.75.

Key Concept

Using completing the square, coordinate translations, and the point-to-line distance formula to solve circle tangency problems
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Topics

Equations and Graphs of Circles

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