Circles intersecting at two points orthogonally. I am finding the following much harder than it probably is!
If a circle $A$ intersects the circle $B$ at two points orthogonally, then why can't $A$ pass through the centre of B?
 A: The circle $A$ can touch its tangents at only one point each. Draw two tangents of $A$ at the points of intersection of the two circles. Since $A$  touches them only at these points and also that tangents intersect at centre of $B$, $A$  cannot touch the lines again, and so cannot pass through the centre of $B$
A: Say $A$ and $B$ intersect in $M$ (and $N$). Then $\angle A'MB' = 90^{\circ}$ where $A',B'$ are a centers of the given circles. So $MB'$ is a tangent on $A$ at $M$ and thus $A$ can't go through $B'$. 
A: Edit: just saw that samjoe already gave this answer, essentially. I'm keeping it here to show that in fact you need only one orthogonal point of intersection to show that $A$ doesn't pass through the center of $B$.
Draw a radius from the center of $B$ to a point of intersection; since the radius is orthogonal to $B$, and $A$ is orthogonal to $B$, the radius is tangent to $A$. But a tangent can intersect $A$ at only one point.
A: Let $A',B'$ be the centers of the circles with radius $a,b$ resp.
Circles intersect orthogonally, I.e points of intersection $M,N$ lie on the Thales circle above $A'B' :=c.$
In these right triangles $c$ is the hypotenuse, recall:
$c^2 = a^2 +b^2$, hence $c \gt a,b.$
One circle cannot pass through centre of the other.
A: Let $X $ and $Y $ be the centres of $A $  and $ B $, respectively, and let $Z $ be one of the two points where those circles intersect each other. The triangle $\triangle XZY$ is a right-angled triangle with hypotenuse $XY $, thus $XY $ is the longest side of the triangle: $XY\gt XZ $ and, since $XZ $ is the radius of $A $, the point $Y $ lies outside of $A $
