$What does the equation (a - b) (x^{2} + y^{2}) - 2 a b x = 0 b e c o m e i f t h e o r i g i n i s s h i f t e d t o t h e p o i n t (\frac{a b}{a - b}, 0) w i t h o u t r o t a t i o n ?$

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Solution

$S u b s t i t u t i n g x = X + \frac{a b}{a - b}, y = Y + 0 i n t h e g i v e n e q u a t i o n, w e g e t : (a - b) [\begin{matrix} {(X + \frac{a b}{a - b})}^{2} + Y^{2} \end{matrix}] - 2 a b \times [\begin{matrix} X + \frac{a b}{a - b} \end{matrix}] = 0 \Rightarrow (a - b) (X^{2} + \frac{a^{2} b^{2}}{(a - b)^{2}} + \frac{2 a b X}{a - b} + Y^{2}) 2 a b X - \frac{2 a^{2} b^{2}}{a - b} = 0 \Rightarrow (a - b) (X^{2} + Y^{2}) + - \frac{a^{2} b^{2}}{a - b} + 2 a b X - 2 a b X - \frac{2 a^{2} b^{2}}{a - b} = 0 \Rightarrow (a - b) (X^{2} + Y^{2}) + - \frac{a^{2} b^{2}}{a - b} = 0 \Rightarrow (a - b)^{2} (X^{2} + Y^{2}) = a^{2} b^{2} Hence,the transformed equation is (a - b)^{2} (X^{2} + Y^{2}) = a^{2} b^{2}$

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