AP EAMCET · Maths · Circle
If \(A\) and \(B\) are the centres of similitude with respect to the circles \(x^2+y^2-14 x+6 y+33=0\) and \(x^2+y^2+30 x-2 y\) \(+1=0\), then midpoint of AB is
- A \(\left(\frac{7}{3}, \frac{4}{5}\right)\)
- B \(\left(\frac{3}{2}, \frac{1}{5}\right)\)
- C \(\left(\frac{39}{2}, \frac{-7}{4}\right)\)
- D \(\left(\frac{39}{4}, \frac{-7}{2}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{39}{4}, \frac{-7}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & S_1 \equiv x^2+y^2-14 x+6 y+33=0 \\ & C_1=(7,-3), r_1=\sqrt{49+3-33}=5 \\ & S_2 \equiv x^2+y^2+30 x-2 y+1=0 \\ & C_2=(-15,1), r_2=\sqrt{225+1-1}=15 \end{aligned}\) Let \(A\) divides \(C_1 C_2\) in \(r_1: r_2\) internally…
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