TS EAMCET · Maths · Circle
The point \((3,-4)\) lies on both the circles \(x^2+y^2-2 x+8 y+13=0 \quad\) and \(x^2+y^2-4 x+6 y+11=0\). Then, the angle between the circles is
- A \(60^{\circ}\)
- B \(\tan ^{-1}\left(\frac{1}{2}\right)\)
- C \(\tan ^{-1}\left(\frac{3}{5}\right)\)
- D \(135^{\circ}\)
Answer & Solution
Correct Answer
(D) \(135^{\circ}\)
Step-by-step Solution
Detailed explanation
Given circles are \(x^2+y^2-2 x+8 y+13=0\) and \(x^2+y^2-4 x+6 y+11=0\). Here, \(C_1=(1,-4), C_2=(2,-3)\), \(\begin{array}{lll} \Rightarrow & & r_1=\sqrt{1+16-13}=2 \\ \text { and } & & r_2=\sqrt{4+9-11}=\sqrt{2} \end{array}\) Now, \(d=C_1 C_2=\sqrt{(2-1)^2+(-3+4)^2}=\sqrt{2}\)…
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