The points of intersection of the circles x^2 | vedclass

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(B) Given equations of circles are:$x^2 + y^2 = 2ax$ ... $(i)$$x^2 + y^2 = 2by$ ... $(ii)$Subtracting $(ii)$ from $(i)$,we get:$2ax - 2by = 0 \implies

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

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(B) Given equations of circles are:$x^2 + y^2 = 2ax$ ... $(i)$$x^2 + y^2 = 2by$ ... $(ii)$Subtracting $(ii)$ from $(i)$,we get:$2ax - 2by = 0 \implies ax = by \implies y = \frac{a}{b}x$Substituting $y = \frac{a}{b}x$ into $(i)$:$x^2 + \left( \frac{a}{b}x \right)^2 = 2ax$$x^2 + \frac{a^2}{b^2}x^2 = 2ax$$x^2 \left( 1 + \frac{a^2}{b^2} \right) = 2ax$$x^2 \left( \frac{a^2 + b^2}{b^2} \right) = 2ax$$x \left[ x \left( \frac{a^2 + b^2}{b^2} \right) - 2a \right] = 0$This gives $x = 0$ or $x = \frac{2ab^2}{a^2 + b^2}$.If $x = 0$,then $y = \frac{a}{b}(0) = 0$.If $x = \frac{2ab^2}{a^2 + b^2}$,then $y = \frac{a}{b} \left( \frac{2ab^2}{a^2 + b^2} \right) = \frac{2a^2b}{a^2 + b^2}$.Thus,the points of intersection are $(0, 0)$ and $\left( \frac{2ab^2}{a^2 + b^2}, \frac{2a^2b}{a^2 + b^2} \right)$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

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

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