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