A circle C₁ of radius r untis rolls outside the circle ${\mathrm{C}}_2:{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{rx}=0$ touching it externally. The line of centers has an inclination 60°. Then

Detailed Solution

Step 1: Find the Center and Radius of C₂

The equation x² + y² + 2rx = 0 can be rewritten as x² + y² + 2rx = 0.

Compare with the general form x² + y² + 2gx + 2fy + c = 0:

• g = r
• f = 0
• c = 0

Center of C₂ is (−g, −f) = (−r, 0).

Radius of C₂ is √(g² + f² − c) = √(r² + 0 − 0) = r.

Result: The center is at (−r, 0) and the radius is r.

Step 2: Center of C₁

Let the center of the rolling circle C₁ be (h, k). Since C₁ touches C₂ externally, the distance between their centers will be equal to r₁ + r₂ = r + r = 2r.

Let (h, k) be located such that the line joining the centers has 60° inclination with the x-axis:

• h + r = 2r cos(60°) = 2r × 1/2 = r
• k − 0 = 2r sin(60°) = 2r × √3/2 = r√3

Since the center of C₂ is at (−r, 0), if the center of C₁ is (h, k), the vector joining (−r, 0) to (h, k) has length 2r and inclination 60°:

• h = −r + 2r cos(60°) = −r + r = 0
• k = 0 + 2r sin(60°) = 0 + r√3 = r√3

Step 3: The Locus/Equation of Center of C₁

If the circle rolls, the locus of the center of C₁ will also be a circle of radius 2r centered at (−r, 0):

Equation: (x + r)² + y² = (2r)²

Step 4: Final Answers

• The center of the rolling circle C₁: (0, r√3)
• The locus: (x + r)² + y² = 4r²
• At inclination 60°, the coordinates of the center: (0, r√3)