Question

If A and B are the points of intersection of the circles \( x^2 + y^2 - 4x + 6y - 3 = 0 \) and \( x^2 + y^2 + 2x - 2y - 2 = 0 \), then the distance between A and B is:

• A. \( \frac{13}{10} \)
• B. \( \frac{\sqrt{41}}{3} \)
• C. \( \frac{\sqrt{231}}{5} \)
• D. \( \frac{26}{5} \)

Hint:

To find the distance between the intersection points of two circles, use the geometry of the system of equations and apply the distance formula.

Answer 1

The Correct Option is C

Step 1: Equations of the circles.
The equations of the two circles are: \[ x^2 + y^2 - 4x + 6y - 3 = 0 \] and \[ x^2 + y^2 + 2x - 2y - 2 = 0 \]
Step 2: Subtract the two equations.
Subtract the second equation from the first: \[ (x^2 + y^2 - 4x + 6y - 3) - (x^2 + y^2 + 2x - 2y - 2) = 0 \] Simplifying: \[ -6x + 8y - 1 = 0 \quad \Rightarrow \quad 3x - 4y = \frac{1}{2} \]
Step 3: Find the distance between the points.
Using the distance formula between the points of intersection of two circles, we find the distance between A and B to be: \[ \text{Distance} = \frac{\sqrt{231}}{5} \]