Examkin

Question

Difficulty: Very hardEquations of Circles in the Coordinate Plane

In the xyxy-plane, the line y=mxy = mx, where mm is a positive constant, is tangent to the circle defined by the equation x2+y2−6x−8y+16=0x^2 + y^2 - 6x - 8y + 16 = 0. What is the value of mm?

  1. 724\frac{7}{24}Answer
  2. B
    247\frac{24}{7}
  3. C
    34\frac{3}{4}
  4. D
    712\frac{7}{12}

Answer

724\frac{7}{24}
The correct answer is 724\frac{7}{24}. Standardizing the circle equation gives (x−3)2+(y−4)2=9(x-3)^2 + (y-4)^2 = 9, identifying the center as (3,4)(3, 4) and the radius as 33. A line mx−y=0mx - y = 0 is tangent to the circle if the distance from (3,4)(3, 4) to the line is 33. Applying the distance formula yields ∣3m−4∣m2+1=3\frac{|3m - 4|}{\sqrt{m^2 + 1}} = 3, which simplifies to m=724m = \frac{7}{24} after squaring and solving for mm.

Step-by-Step Solution

1
Complete the square for the circle's equation to find the center and radius.
The given equation x2+y2−6x−8y+16=0x^2 + y^2 - 6x - 8y + 16 = 0 can be rewritten as (x−3)2−9+(y−4)2−16+16=0(x-3)^2 - 9 + (y-4)^2 - 16 + 16 = 0, which simplifies to (x−3)2+(y−4)2=9(x-3)^2 + (y-4)^2 = 9. Thus, the circle has center (3,4)(3, 4) and radius r=9=3r = \sqrt{9} = 3.
Converting the general form equation of a circle to standard form is necessary to find the coordinates of its center and its radius.
2
Set up the equation for the distance from the center of the circle to the tangent line.
The line is given by y=mxy = mx, which can be written in standard form as mx−y=0mx - y = 0. The perpendicular distance from the center (3,4)(3, 4) to this line must equal the circle's radius 33. Using the point-to-line distance formula: ∣m(3)−(4)∣m2+(−1)2=3\frac{|m(3) - (4)|}{\sqrt{m^2 + (-1)^2}} = 3.
A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius.
3
Solve the distance equation for the constant mm.
∣3m−4∣=3m2+1|3m - 4| = 3\sqrt{m^2 + 1}. Squaring both sides gives (3m−4)2=9(m2+1)⇒9m2−24m+16=9m2+9(3m - 4)^2 = 9(m^2 + 1) \Rightarrow 9m^2 - 24m + 16 = 9m^2 + 9. Subtracting 9m29m^2 from both sides gives −24m+16=9⇒24m=7⇒m=724-24m + 16 = 9 \Rightarrow 24m = 7 \Rightarrow m = \frac{7}{24}.
Squaring both sides eliminates the absolute value and the radical, allowing us to isolate and solve for mm algebraically.

Key Concept

Equations of circles and the relationship between a circle and its tangent lines in the coordinate plane

Alternative Method

The problem can also be solved using geometry and right-triangle trigonometry. The distance from the origin O(0,0)O(0,0) to the center C(3,4)C(3,4) is 55. The radius to the point of tangency TT is 33, forming a right triangle OTCOTC with hypotenuse OC=5OC = 5 and leg CT=3CT = 3. The other leg is OT=4OT = 4. The angle θ\theta that OCOC makes with the positive xx-axis has tan⁡θ=43\tan\theta = \frac{4}{3}, and the angle α\alpha between OCOC and OTOT has tan⁡α=34\tan\alpha = \frac{3}{4}. The slope of the tangent line OTOT is m=tan⁡(θ−α)m = \tan(\theta - \alpha). Applying the tangent subtraction formula tan⁡(θ−α)=tan⁡θ−tan⁡α1+tan⁡θtan⁡α\tan(\theta - \alpha) = \frac{\tan\theta - \tan\alpha}{1 + \tan\theta\tan\alpha} gives m=724m = \frac{7}{24}.
Estimated Time:3m 0s
Rate this question

Topics

Equations of Circles in the Coordinate Plane

More questions from this topic