Question

The equation of the circle of radius 3 units which touches the circles $ x^2 + y^2 - 6|x| = 0 $ is:

• A. \( x^2 + y^2 + 6\sqrt{3}y - 18 = 0 \) or \( x^2 + y^2 - 6\sqrt{3}y - 18 = 0 \)
• B. \( x^2 + y^2 + 4\sqrt{3}y + 18 = 0 \) or \( x^2 + y^2 - 4\sqrt{3}y + 18 = 0 \)
• C. \( x^2 + y^2 + 6\sqrt{3}y + 18 = 0 \) or \( x^2 + y^2 - 6\sqrt{3}y + 18 = 0 \)
• D. \( x^2 + y^2 + 4\sqrt{3}y - 18 = 0 \) or \( x^2 + y^2 - 4\sqrt{3}y - 18 = 0 \)

Hint:

When dealing with the tangency of two circles, use the geometric properties of the circles and their relative positions to solve for the equation of the new circle.

Answer 1

The Correct Option is C

The equation of the given circle \( x^2 + y^2 - 6|x| = 0 \) represents two circles, one with \( x \geq 0 \) and another with \( x < 0 \). 
This equation can be rewritten as: \[ x^2 + y^2 = 6x \quad \text{for} \quad x \geq 0. \] Now, to find the equation of the circle that touches these two circles and has a radius of 3, we use the general formula for the equation of a circle: \[ (x - h)^2 + (y - k)^2 = r^2, \] where \( (h, k) \) is the center of the circle and \( r \) is its radius. By solving this geometrically or using the condition for tangency, we find that the correct equation is \( x^2 + y^2 + 6\sqrt{3}y + 18 = 0 \) or \( x^2 + y^2 - 6\sqrt{3}y + 18 = 0 \).