The points of intersection of circles {x^2} + | vedclass

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Question

(b) Given circles are ${x^2} + {y^2} = 2ax$$&hellip;..(i)$

and ${x^2} + {y^2} = 2by$$&hellip;..(ii)$

$(i) -(ii)$ $&rArr;$ $0 = 2(ax - by)$ $&rAr

Vedclass · Accepted Answer

$(0, 0)$, $\left( {\frac{{2a{b^2}}}{{{a^2} + {b^2}}},\frac{{2b{a^2}}}{{{a^2} + {b^2}}}} \right)$

Vedclass · Answer

(b) Given circles are ${x^2} + {y^2} = 2ax$$&hellip;..(i)$

and ${x^2} + {y^2} = 2by$$&hellip;..(ii)$

$(i) -(ii)$ $&rArr;$ $0 = 2(ax - by)$ $&rArr;$ $y = \frac{a}{b}x$

From $(i),$ ${x^2} + \frac{{{a^2}}}{{{b^2}}}{x^2} = 2ax$

$&rArr;$ $x\left\{ {\left( {1 + \frac{{{a^2}}}{{{b^2}}}} \right)x - 2a} \right\} = 0$

$&rArr;$ $x = 0,\;\frac{{2a{b^2}}}{{{a^2} + {b^2}}}$

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 $\left( {\frac{{2a{b^2}}}{{{a^2} + {b^2}}},\;\frac{{2{a^2}b}}{{{a^2} + {b^2}}}} \right)$.

The points of intersection of circles ${x^2} + {y^2} = 2ax$ and ${x^2} + {y^2} = 2by$ are

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