Question

If a direct common tangent is drawn to the circles \( x^2 + y^2 - 6x + 4y + 9 = 0 \) and \( x^2 + y^2 + 2x - 2y + 1 = 0 \) that touches the circles at points \( A \) and \( B \), then \( AB = \):

• A. 9
• B. 16
• C. \( \sqrt{6} \)
• D. \( 2\sqrt{6} \)

Hint:

To find the length of the direct common tangent between two circles, use the formula \( AB = \sqrt{d^2 - (r_1 - r_2)^2} \), where \( d \) is the distance between the centers and \( r_1 \) and \( r_2 \) are the radii of the circles.

Answer 1

The Correct Option is D

We are given the equations of two circles:
1. \( x^2 + y^2 - 6x + 4y + 9 = 0 \)
2. \( x^2 + y^2 + 2x - 2y + 1 = 0 \)
We are asked to find the distance between the points of tangency \( A \) and \( B \), where a direct common tangent touches both circles.
Step 1: First, rewrite both equations of the circles in standard form by completing the square for both \( x \) and \( y \).
For the first circle \( x^2 + y^2 - 6x + 4y + 9 = 0 \):
- Complete the square for \( x \): \( x^2 - 6x \rightarrow (x - 3)^2 - 9 \)
- Complete the square for \( y \): \( y^2 + 4y \rightarrow (y + 2)^2 - 4 \)
Thus, the equation becomes: \[ (x - 3)^2 + (y + 2)^2 = 4 \] So, the center is \( (3, -2) \) and the radius is \( 2 \). For the second circle \( x^2 + y^2 + 2x - 2y + 1 = 0 \):
- Complete the square for \( x \): \( x^2 + 2x \rightarrow (x + 1)^2 - 1 \)
- Complete the square for \( y \): \( y^2 - 2y \rightarrow (y - 1)^2 - 1 \)
Thus, the equation becomes: \[ (x + 1)^2 + (y - 1)^2 = 1 \] So, the center is \( (-1, 1) \) and the radius is \( 1 \). 
Step 2: Next, we use the formula for the distance between two points on the direct common tangent. The distance \( AB \) between the points of tangency of the direct common tangent is given by: \[ AB = \sqrt{d^2 - (r_1 - r_2)^2} \] where: - \( d \) is the distance between the centers of the two circles
- \( r_1 \) and \( r_2 \) are the radii of the two circles
The distance between the centers is: \[ d = \sqrt{(3 - (-1))^2 + (-2 - 1)^2} = \sqrt{(3 + 1)^2 + (-2 - 1)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] The radii of the two circles are \( r_1 = 2 \) and \( r_2 = 1 \). So, the distance between the points of tangency is: \[ AB = \sqrt{5^2 - (2 - 1)^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6} \] Thus, the distance \( AB \) is \( 2\sqrt{6} \).