Question

The points of intersection of circles $ (x + 1)^2 + y^2 = 4 \quad \text{and} \quad (x - 1)^2 + y^2 = 9 \quad \text{are} \quad (a, \pm b), \quad \text{then} \quad (a, b) \quad \text{equals to} $

• A. \( \left(1.25, \frac{3}{4} \sqrt{7}\right) \)
• B. \( \left(-1.25, \frac{3}{4} \sqrt{7}\right) \)
• C. \( (-1, 2) \)
• D. \( (1, 3) \)

Hint:

To find the points of intersection of two circles, subtract the equations and use the difference of squares to solve for \( x \), then substitute into one of the circle equations to find \( y \).

Answer 1

The Correct Option is B

We are given two equations of circles: 1. \( (x + 1)^2 + y^2 = 4 \) 2. \( (x - 1)^2 + y^2 = 9 \) To find the points of intersection, subtract the second equation from the first: \[ \left[ (x + 1)^2 + y^2 \right] - \left[ (x - 1)^2 + y^2 \right] = 4 - 9 \] Simplifying: \[ (x + 1)^2 - (x - 1)^2 = -5 \] Using the difference of squares formula: \[ \left[ (x + 1) - (x - 1) \right] \cdot \left[ (x + 1) + (x - 1) \right] = -5 \] \[ (2) \cdot (2x) = -5 \] \[ 4x = -5 \] \[ x = -\frac{5}{4} = -1.25 \] Now, substitute \( x = -1.25 \) into either of the original circle equations. Using the first equation: \[ (-1.25 + 1)^2 + y^2 = 4 \] \[ (-0.25)^2 + y^2 = 4 \] \[ 0.0625 + y^2 = 4 \] \[ y^2 = 4 - 0.0625 = 3.9375 \] \[ y = \pm \frac{3}{4} \sqrt{7} \] Thus, the points of intersection are \( \left(-1.25, \pm \frac{3}{4} \sqrt{7}\right) \).