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If radius of the director circle of the ellipse $\frac{(3 x+4 y-2)^2}{100}+\frac{(4 x-3 y+5)^2}{625}=1$ is
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$\sqrt{29}$
$\begin{aligned} & \frac{(3 x+4 y-2)^2}{100}+\frac{(4 x-3 y+5)^2}{625}=1 \\ & \Rightarrow \frac{\left(\frac{3 x+4 y-2}{5}\right)^2}{4}+\frac{\left(\frac{4 x-3 y+5}{5}\right)^2}{25}=1 \\ & \Rightarrow a^2=4 \text { and } b^2=25\end{aligned}$
Radius of director circle $=\sqrt{a^2+b^2}=\sqrt{29}$
Radius of director circle $=\sqrt{a^2+b^2}=\sqrt{29}$
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