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AP EAMCET · Maths · Circle

If the angle between the circles \(x^2+y^2-2 x-4 y+c=0\) and \(x^2+y^2-4 x-2 y+4=0\) is \(60^{\circ}\), then \(c\) is equal to

  1. A \(\frac{3 \pm \sqrt{5}}{2}\)
  2. B \(\frac{6 \pm \sqrt{5}}{2}\)
  3. C \(\frac{9 \pm \sqrt{5}}{2}\)
  4. D \(\frac{7 \pm \sqrt{5}}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{7 \pm \sqrt{5}}{2}\)

Step-by-step Solution

Detailed explanation

We have, \(\begin{aligned} & \mathrm{C}_1: x^2+y^2-2 x-4 y+c=0 ... (i)\\ & \mathrm{C}_2: x^2+y^2-4 x-2 y+4=0... (ii) \end{aligned}\) So, centre of circle \(\mathrm{C}_1=(1,2), r_1=\sqrt{5-c}\) and centre of circle \(\mathrm{C}_2=(2,1), r_2=1\) Now, angle between the two circles,…