Question

The slope of a common tangent to the circles \(x^2 + y^2 = 16\) and \((x-9)^2 + y^2 = 16\) is

• A. \(\frac{8}{\sqrt{13}}\)
• B. \(\frac{4}{\sqrt{13}}\)
• C. \(\frac{\sqrt{17}}{8}\)
• D. \(\frac{8}{\sqrt{17}}\)

Hint:

For two circles with equal radii, the internal tangents pass through the midpoint of the centers. This simplifies the problem significantly compared to the general case where they divide the line of centers in ratio \(r_1:r_2\).

Answer 1

The Correct Option is D

Step 1: Analyze Circles:
Circle 1: Center \(C_1(0,0)\), Radius \(r_1 = 4\). Circle 2: Center \(C_2(9,0)\), Radius \(r_2 = 4\). Since radii are equal, the external common tangents are parallel to the line of centers (x-axis), so their slope is 0. The internal common tangents intersect at the midpoint of the centers.
Step 2: Find Intersection of Internal Tangents:
Midpoint \(M = \frac{C_1 + C_2}{2} = (\frac{9}{2}, 0)\). The tangent passes through \(M(\frac{9}{2}, 0)\).
Step 3: Equation of Tangent:
Let slope be \(m\). Equation: \(y - 0 = m(x - \frac{9}{2}) \implies mx - y - \frac{9m}{2} = 0 \implies 2mx - 2y - 9m = 0\).
Step 4: Distance Condition:
Distance from \(C_1(0,0)\) to the tangent must be equal to radius \(r_1 = 4\). \[ \frac{|2m(0) - 2(0) - 9m|}{\sqrt{(2m)^2 + (-2)^2}} = 4 \] \[ \frac{|-9m|}{\sqrt{4m^2 + 4}} = 4 \] Square both sides: \[ \frac{81m^2}{4(m^2+1)} = 16 \] \[ 81m^2 = 64(m^2 + 1) \] \[ 81m^2 - 64m^2 = 64 \] \[ 17m^2 = 64 \implies m^2 = \frac{64}{17} \implies m = \pm \frac{8}{\sqrt{17}} \] Final Answer:
The slope is \(\frac{8}{\sqrt{17}}\).