Question

Find the interval of \(\lambda\) for which exactly two common tangents can be drawn to \(x^2+y^2-4x-4y+6=0\) and \(x^2+y^2-10x-10y+\lambda=0\).

• A. (12, 24)
• B. (12, 32)
• C. (18, 42)
• D. (18, 48)

Hint:

Always verify that the radius remains a real number (\(\lambda < 50\)) when finding the interval for \(\lambda\).

Answer 1

The Correct Option is C

Concept: Two circles have exactly two common tangents when they intersect at two distinct points. This is equivalent to the condition that the distance between centers \(d\) satisfies \(|r_1 - r_2| < d < r_1 + r_2\).

Step 1: Determine the center and radius for both circles.
Circle 1: \(x^2+y^2-4x-4y+6=0\). Center \(C_1 = (2, 2)\), \(r_1^2 = 2^2+2^2-6 = 2 \implies r_1 = \sqrt{2}\).
Circle 2: \(x^2+y^2-10x-10y+\lambda=0\). Center \(C_2 = (5, 5)\), \(r_2^2 = 5^2+5^2-\lambda = 50-\lambda \implies r_2 = \sqrt{50-\lambda}\).
Distance between centers \(d = \sqrt{(5-2)^2 + (5-2)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}\).

Step 2: Set up the inequality for two intersection points.
\(|r_1 - r_2| < d < r_1 + r_2\) \[ |\sqrt{50-\lambda} - \sqrt{2}| < 3\sqrt{2} < \sqrt{50-\lambda} + \sqrt{2} \]

Step 3: Solve the inequality part 1: \(d < r_1 + r_2\).
\(3\sqrt{2} < \sqrt{50-\lambda} + \sqrt{2} \implies 2\sqrt{2} < \sqrt{50-\lambda} \implies 8 < 50-\lambda \implies \lambda < 42\).
Actually, solving \(d < r_1+r_2\) yields \(\lambda < 32\).

Step 4: Solve the inequality part 2: \(d > |r_1 - r_2|\).
\(3\sqrt{2} > \sqrt{50-\lambda} - \sqrt{2} \implies 4\sqrt{2} > \sqrt{50-\lambda} \implies 32 > 50-\lambda \implies \lambda > 18\). center minipage0.3

Interval: (18, 42) minipage center