AP EAMCET · Maths · Circle
The distance of the origin from the external centre of similitude for the circles \(x^2+y^2-8 x-10 y-8=0\) and \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-2 \mathrm{y}-2=0\) is
- A \(\frac{3 \sqrt{26}}{5}\)
- B \(\frac{\sqrt{290}}{9}\)
- C \(\frac{\sqrt{290}}{5}\)
- D \(\frac{\sqrt{26}}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \sqrt{26}}{5}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{S}_1: x^2+y^2-8 x-10 y-8=0\) \(\begin{aligned} & \mathrm{C}_1=(4,5) \& r_1=\sqrt{16+25+8}=7 \\ & \text { and } \mathrm{S}_2: x^2+y^2+2 x-2 y-2=0 \\ & \mathrm{C}_2=(-1,1) \& r_2=\sqrt{1+1+2}=2\end{aligned}\) Using section formula, external centre of similitude is:…
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