Circles intersecting at two points orthogonally.

Question

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?

Answer 1

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 $B$

Answer 2

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 $B'$.

Answer 3

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.

Answer 4

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.

Answer 5

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 $