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The points of intersection of circles ${x^2} + {y^2} = 2ax$ and ${x^2} + {y^2} = 2by$ are
$(0, 0)$, $(a, b)$
$(0, 0)$, $\left( {\frac{{2a{b^2}}}{{{a^2} + {b^2}}},\frac{{2b{a^2}}}{{{a^2} + {b^2}}}} \right)$
$(0, 0)$, $\left( {\frac{{{a^2} + {b^2}}}{{{a^2}}},\frac{{{a^2} + {b^2}}}{{{b^2}}}} \right)$
None of the above
Solution
(b) Given circles are ${x^2} + {y^2} = 2ax$$…..(i)$
and ${x^2} + {y^2} = 2by$$…..(ii)$
$(i) -(ii)$ $⇒$ $0 = 2(ax – by)$ $⇒$ $y = \frac{a}{b}x$
From $(i),$ ${x^2} + \frac{{{a^2}}}{{{b^2}}}{x^2} = 2ax$
$⇒$ $x\left\{ {\left( {1 + \frac{{{a^2}}}{{{b^2}}}} \right)x – 2a} \right\} = 0$
$⇒$ $x = 0,\;\frac{{2a{b^2}}}{{{a^2} + {b^2}}}$
For $x = 0$, $y = 0$ and for $x = \frac{{2a{b^2}}}{{{a^2} + {b^2}}}$, $y = \frac{{2{a^2}b}}{{{a^2} + {b^2}}}$
The points of intersection are $(0, 0)$ and $\left( {\frac{{2a{b^2}}}{{{a^2} + {b^2}}},\;\frac{{2{a^2}b}}{{{a^2} + {b^2}}}} \right)$.
