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COMEDK · Maths · 12. Circle

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

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

Answer & Solution

Correct Answer

(B) \(\left(-1.25, \frac{3}{4} \sqrt{7}\right)\)

Step-by-step Solution

Detailed explanation

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…