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(B) Let the centres of the two circles be $A$ and $B$. Since the radii are equal,let the radius be $r$. The circles intersect at $D(0, 1)$ and $E(0, -

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$2$

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(B) Let the centres of the two circles be $A$ and $B$. Since the radii are equal,let the radius be $r$. The circles intersect at $D(0, 1)$ and $E(0, -1)$.Let the centre of the first circle be $A(-h, 0)$ and the second be $B(h, 0)$.The equation of the circle with centre $A(-h, 0)$ is $(x+h)^2 + y^2 = r^2$. Since it passes through $(0, 1)$,we have $h^2 + 1 = r^2$.The tangent at $(0, 1)$ to this circle has the equation $x(0+h) + y(1) = r^2$,which simplifies to $hx + y = r^2$.This tangent passes through the centre of the other circle $B(h, 0)$,so $h(h) + 0 = r^2$,which means $h^2 = r^2$.Substituting $r^2 = h^2$ into $h^2 + 1 = r^2$ gives $h^2 + 1 = h^2$,which is impossible. Let's re-evaluate.The tangent at $(0, 1)$ to the circle with centre $A$ is perpendicular to the radius $AD$. The slope of $AD$ is $\frac{1-0}{0-(-h)} = \frac{1}{h}$. Thus,the slope of the tangent is $-h$.The equation of the tangent is $y - 1 = -h(x - 0)$,or $y = -hx + 1$.This line passes through the centre of the other circle $B(h, 0)$,so $0 = -h(h) + 1$,which gives $h^2 = 1$,so $h = 1$.The distance between the centres $A(-1, 0)$ and $B(1, 0)$ is $1 - (-1) = 2$.

Two circles with equal radii intersect at the points $(0, 1)$ and $(0, -1).$ The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is

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