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

Answer & Solution

Correct Answer

(D) \(\left(-1.25, \dfrac{3}{4} \sqrt{7}\right)\)

Step-by-step Solution

Detailed explanation

The given equations of the circles are: \((x+1)^2 + y^2 = 4\) (1) \((x-1)^2 + y^2 = 9\) (2) Expanding both equations: \(x^2 + 2x + 1 + y^2 = 4 \Rightarrow x^2 + y^2 + 2x = 3\) (3) \(x^2 - 2x + 1 + y^2 = 9 \Rightarrow x^2 + y^2 - 2x = 8\) (4) Subtracting equation (4) from…
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