#5 Inversion I: First Contact

We have \(OA \cdot OA'=8^2=64\). Thus \(5\cdot OA'=64\), so \(OA'=\frac{64}{5}\).

For the image \(P'\), \(OP\cdot OP'=R^2\). If \(OP=R\), then \(R\cdot OP'=R^2\), so \(OP'=R=OP\). Point \(P'\) lies on ray \(OP\), hence \(P'=P\).

We get \(R^2=4\cdot 25=100\), so \(R=10\). Since the points lie on one ray, the inversion condition is fully satisfied.

If \(P\) lies on the circle, then \(OP=3\). Then \(3\cdot OP'=144\), so \(OP'=48\). The image is the circle with centre \(O\) and radius \(48\).

If \(P\) lies on a line through \(O\), then ray \(OP\) lies on the same line. Therefore \(P'\) also lies on this line. This holds for every point, and inversion is its own inverse, so the whole line maps to itself.

We have \(R^2=7\cdot 28=196\), hence \(R=14\). The inversion with centre \(O\) and radius \(14\) maps these points to each other.

Let \(P'\) be the image of \(P\). Since \(OP\cdot OP'=OH\cdot OH'=R^2\), we have \(\frac{OP}{OH'}=\frac{OH}{OP'}\). The angle at \(O\) is common, so \(\triangle OPH\sim\triangle OH'P'\). From \(\angle OHP=90^\circ\), it follows that \(\angle OP'H'=90^\circ\). Thus \(P'\) lies on the circle with diameter \(OH'\).

Here \(OH=5\), so \(OH'=\frac{100}{5}=20\). The image of the line is the circle with diameter \(OH'\), hence its radius is \(10\).

Since \(\omega\) passes through \(O\), its image is a line not passing through \(O\). All points \(A',B',C'\) lie on this image, hence they are collinear.

A line not passing through the centre maps to a circle through the centre. Hence the preimages of \(A',B',C'\) lie on one circle with point \(O\). Therefore \(O,A,B,C\) are concyclic.

The circle passes through \(O\), so its image is a line. For \(Q'\), \(OQ\cdot OQ'=36\), hence \(OQ'=\frac{36}{8}=\frac{9}{2}\). The image line passes through \(Q'\) perpendicular to \(OQ\), so its distance from \(O\) is \(\frac{9}{2}\).

Since \(l\) is tangent to the circle of inversion at \(T\), we have \(OT\perp l\) and \(OT=R\). Thus \(T\) is fixed. By the line-image lemma, the image of \(l\) is the circle with diameter \(OT'\), that is, with diameter \(OT\).

Point \(P\) is not the centre of inversion, so the angle between the curves is preserved. Therefore the angle between the images at \(P'\) equals the original angle \(40^\circ\).

At \(T\), the angle between the circles is \(0^\circ\). Inversion preserves angles at points different from the centre, so the angle between the images at \(T'\) is also \(0^\circ\). Hence the images are tangent.

Both circles pass through the centre of inversion, so both map to lines. Their common point \(A\) maps to \(A'\), so \(A'\) lies on both image lines. Therefore the lines meet at \(A'\).

Orthogonality gives \(OC^2=R^2+r^2\). For a secant \(OAB\), by the power of point \(O\) with respect to \(\omega\), \(OA\cdot OB=OC^2-r^2=R^2\). Thus \(A\) and \(B\) are inverse points. Every secant through \(O\) swaps points of the circle in pairs, so \(\omega\) maps to itself.

Point \(Q\) maps to \(Q'\), where \(OQ\cdot OQ'=R^2\). Since \(OQ=2r\), we have \(OQ'=\frac{R^2}{2r}\). A circle through \(O\) maps to a line; in this configuration, that line passes through \(Q'\) perpendicular to \(OC\). Therefore the distance from \(O\) to the image line is \(\frac{R^2}{2r}\).

Since \(\omega\) passes through the centre of inversion, its image is a line. Point \(A\) lies on \(\omega\), and its image by the condition is \(B\). Therefore the image line passes through \(B\).

A circle through \(O\) maps to a line perpendicular to the line joining \(O\) with its centre. If two circles are tangent at \(O\), their centres lie on one line perpendicular to the common tangent. Therefore the two image lines are perpendicular to the same line. Hence they are parallel.

The images of \(l\) and \(m\) are circles through \(O\). Since \(P\in l\cap m\), point \(P'\) lies on both image circles. Inversion preserves the angle at \(P\ne O\), so the angle between the circles at \(P'\) equals the angle between lines \(l\) and \(m\).

For the circle to be invariant, for every secant through \(O\) the product of distances to the intersection points must equal \(R^2\). This product is the power of the point: \(OC^2-r^2=13^2-5^2=169-25=144\). Hence \(R=12\).

At \(T\), the angle between \(l\) and \(\omega\) is \(0^\circ\). Since \(T\ne O\), inversion preserves this angle. Therefore the angle between the images at \(T'\) is also \(0^\circ\), so the image of \(l\) is tangent to the image of \(\omega\).

Since both circles pass through \(O\), their images are lines. Point \(A\) maps to \(A'\), and both image lines pass through \(A'\). Inversion preserves the angle between curves at \(A\ne O\), so the angle between the image lines equals the original angle between the circles.

Circles \(\omega_1,\omega_2,\omega_3\) pass through the centre of inversion, so their images are three lines \(l_1,l_2,l_3\). Point \(A\) lies on \(\omega_1\) and \(\omega_2\), hence \(A'\) lies on \(l_1\) and \(l_2\). Similarly, \(B'=l_2\cap l_3\), and \(C'=l_3\cap l_1\). Thus the configuration of three circles through \(O\) becomes a triangle formed by three lines.