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

• A

  $\left(1.25, \frac{3}{4} \sqrt{7}\right)$
• Correct answer
  B

  $\left(-1.25, \frac{3}{4} \sqrt{7}\right)$
• C

  $(-1,2)$
• D

  $(1,3)$

Solution

On subtracting Eq. (ii) from Eq. (i), we get $\begin{array}{rlrl} & (x+1)^2-(x-1)^2=4-9 \\ \Rightarrow & \left(x^2+2 x+1\right)-\left(x^2-2 x+1\right)=-5 \\ \Rightarrow & 4 x=-5 \\ \Rightarrow & x=-1.25 \end{array}$ On putting, $x=-1.25$ into Eq. (i), we get $\begin{aligned} & (-0.25)^2+y^2=4 \\ & \Rightarrow \quad y^2=3.9375 \Rightarrow y= \pm \sqrt{3.9375} \\ & \Rightarrow \quad y= \pm \frac{3}{4} \sqrt{7} \\ & \therefore \quad a=-1.25 \text { and } \quad b=\frac{3}{4} \sqrt{7} \\ & \therefore \quad(a, b)=\left(-1.25, \frac{3}{4} \sqrt{7}\right) \\ & \end{aligned}$

Hence, the correct answer is option: 'B'.