Olympiad Geometry

Put the line through $P$ in the form $P + t\mathbf u$ with $|\mathbf u| = 1$. The point is on the circle when $|P - O + t\mathbf u|^2 = r^2$, that is $t^2 + 2t\,\mathbf u\cdot(P - O) + \left(PO^2 - r^2\right) = 0$. The two roots are the signed distances $\overline{PX}$ and $\overline{PY}$, and by Vieta their product is the constant term $PO^2 - r^2$, whatever the direction $\mathbf u$. That is the whole theorem.

For the axis, write each circle as $x^2 + y^2 + D_ix + E_iy + F_i = 0$; the left side at $(x, y)$ is exactly the power of $(x, y)$. Equal power means the difference of two such expressions vanishes, and the $x^2 + y^2$ terms cancel: what is left is linear, a line with normal $(D_1 - D_2, E_1 - E_2)$, parallel to the line of centers. If the circles meet, both common points have power $0$ for both circles, so the axis is the common chord; if they touch, it is the common tangent. A point on two of the three axes has $\operatorname{pow}_1 = \operatorname{pow}_2 = \operatorname{pow}_3$, so it lies on the third. Concurrency is the transitivity of equality.

Step 1 — the powers are $x^2 + y^2 - 25$, $\;x^2 + y^2 - 16x + 55$ and $x^2 + y^2 - 12y + 11$. Setting the first equal to the second gives $-25 = -16x + 55$, so $x = 5$; the first equal to the third gives $-25 = -12y + 11$, so $y = 3$. The radical center is $R = (5, 3)$.

Step 2 — its power is $25 + 9 - 25 = 9$ for all three circles (check $\omega_2$: $9 + 9 - 9 = 9$; $\omega_3$: $25 + 9 - 25 = 9$). So the tangents from $R$ to each circle have length $3$.

Step 3 — the three pairs behave differently. $\omega_1$ and $\omega_2$ have centers $8 = 5 + 3$ apart, so they touch at $(5, 0)$ and their axis $x = 5$ is the common tangent there. $\omega_1$ and $\omega_3$ meet (centers $6$ apart, radii $5, 5$). $\omega_2$ and $\omega_3$ are disjoint (centers $10$ apart, radii sum $8$) — the axis exists anyway, though no picture shows it as a chord.

Step 4 — the circle $\gamma$ with center $R$ and radius $3$ passes through $(5, 0)$. If $T$ is a point where $\gamma$ meets $\omega_i$, then $RT = 3$ is a tangent length from $R$, so $RT$ is tangent to $\omega_i$ at $T$, i.e. $RT \perp O_iT$: the radius of $\gamma$ at $T$ is tangent to $\omega_i$, and the circles cross at right angles.

How to find it. Subtracting circle equations kills the quadratic terms; you never solve a quadratic. The orthogonal circle is the radical center's power made visible.

Step 1 — line $AB$ is the radical axis: $A$ and $B$ have power $0$ with respect to both circles.

Step 2 — $AB$ is perpendicular to the line of centers, while $PQ$ is not (a line perpendicular to the line of centers and tangent to both circles would touch both at points of that line, forcing the circles to be tangent to each other, not to meet twice). So $AB$ and $PQ$ meet at a point $M$.

Step 3 — $M$ is on the radical axis, and its powers are the squares of its tangent lengths: $MP^2 = \operatorname{pow}_1(M) = \operatorname{pow}_2(M) = MQ^2$. Since $P \ne Q$ and $M$ is on line $PQ$ with $MP = MQ$, $M$ is the midpoint. $\blacksquare$ As a bonus, $MP^2 = MA\cdot MB$.

How to find it. A tangent length squared is a power. The moment a problem has a common tangent and a common chord, the radical axis is waiting.

Step 1 — let $H$ be the orthocenter of $ABC$ and $A_1$, $B_1$, $C_1$ the feet of its altitudes. Since $\angle AA_1D = 90^\circ$, $A_1$ lies on the circle with diameter $AD$, and line $AH$ meets that circle at $A$ and $A_1$. So the power of $H$ with respect to it is $\overline{HA}\cdot\overline{HA_1}$. In the same way its powers for the circles on $BE$ and $CF$ are $\overline{HB}\cdot\overline{HB_1}$ and $\overline{HC}\cdot\overline{HC_1}$.

Step 2 — these three products are equal. $\angle AA_1B = \angle AB_1B = 90^\circ$, so $A, B, A_1, B_1$ lie on the circle with diameter $AB$, and $\overline{HA}\cdot\overline{HA_1}$ and $\overline{HB}\cdot\overline{HB_1}$ are both the power of $H$ for that circle. Likewise for $B$ and $C$. Hence $H$ has equal power for all three circles.

Step 3 — the same argument runs in each of the four triangles cut out by the four lines, because in each one the segments $AD$, $BE$, $CF$ join a vertex to a point on the opposite side line: in $AEF$, $A$ is joined to $D \in EF$, $E$ to $B \in AF$, $F$ to $C \in AE$; similarly for $BDF$ and $CDE$. So all four orthocenters have equal power for the three circles.

Step 4 — two distinct points of equal power lie on the radical axis of each pair of circles, so all three pairwise axes are the line through them: the circles are coaxal, with all four orthocenters on the common axis. Each pairwise axis is perpendicular to the corresponding line of centers, so all three centers lie on one line perpendicular to the common axis. The centers are the midpoints of $AD$, $BE$, $CF$. $\blacksquare$

How to find it. A circle on a diameter $AD$ contains every point that sees $AD$ at a right angle — including the foot of the altitude from $A$. That is what links these circles to $H$, and $H$ is the one point whose powers you can compare.

For homothety, $P - X \mapsto k(P - X)$ is a scaling of vectors, so parallelism and ratios survive and $|P'Q'| = |k|\,|PQ|$. A homothety taking circle $(O_1, r_1)$ to $(O_2, r_2)$ must send $O_1 \mapsto O_2$ and scale by $|k| = \tfrac{r_2}{r_1}$; the sign of $k$ gives two choices, and each determines the center on line $O_1O_2$ dividing it in ratio $\pm r_1 : r_2$.

For the spiral center $S$, take $S$ on both circles. Inscribed angles on the chord $SB$ of $(XAB)$ and on the chord $SD$ of $(XCD)$ give, in directed angles, $\angle(AS, AB) = \angle(XS, XB) = \angle(XS, XD) = \angle(CS, CD)$ since $X, B, D$ are collinear; similarly $\angle(BS, BA) = \angle(DS, DC)$. So triangles $SAB$ and $SCD$ are directly similar, which says precisely that the spiral similarity about $S$ with angle $\angle ASC$ and ratio $\tfrac{SC}{SA}$ sends $A \mapsto C$ and $B \mapsto D$. In complex numbers the map is $z \mapsto az + b$ with $a = \tfrac{d - c}{b - a} \ne 1$, whose unique fixed point is $S$.

A homothety sends lines to parallel lines, circles to circles, and scales every length by $|k|$ and every area by $k^2$. Locating its center — a centroid, a tangency point, an intersection of common tangents — is the whole problem; the ratio then reads off any pair of corresponding lengths.

Step 1 — the homothety $h$ with center $T$ and ratio $\tfrac52$ maps $\omega$ to $\Omega$ (it fixes $T$, sends the center $(0, -3)$ of $\omega$ to $(0, 0)$, and scales the radius $2$ to $5$). It sends $P$ to the point of $\Omega$ on ray $TP$, which is $M$.

Step 2 — $h$ maps the tangent line to $\omega$ at $P$ (the chord's line) to a parallel line tangent to $\Omega$ at $M$. The tangent to $\Omega$ parallel to a chord touches at a midpoint of one of the two arcs, and because $h$ has positive ratio the tangent at $M$ lies on the side of the chord away from $T$. So $M$ is the midpoint of the arc away from $T$.

Step 3 — for an endpoint $A$ of the chord, the arcs $MA$ and $MB$ are equal, so $\angle MAB = \angle MTA$ (inscribed angles on equal arcs). Triangles $MAP$ and $MTA$ share the angle at $M$, hence are similar, and $\tfrac{MA}{MP} = \tfrac{MT}{MA}$, i.e. $MA^2 = MP\cdot MT$.

Step 4 — numbers. The tangency point is $P = (0, -3) + 2\cdot\left(\tfrac35, \tfrac45\right) = \left(\tfrac65, -\tfrac75\right)$ and $M = T + \tfrac52(P - T) = (3, 4)$, which is on $\Omega$ and on the normal direction $(3, 4)$ of the chord, as Step 2 promised. $MP = \tfrac95\sqrt{10}$, $MT = 3\sqrt{10}$, so $MA^2 = \tfrac{27}{5}\cdot 10 = 54$ and $MA = 3\sqrt6$. Directly: $M$ is $\tfrac{27}{5}$ from the chord and the half-chord is $\sqrt{25 - \tfrac{4}{25}}$, so $MA^2 = \tfrac{729}{25} + \tfrac{621}{25} = 54$.

How to find it. Two tangent circles plus a line tangent to the small one: the homothety at the tangency point is the first thing to try, and it turns the tangent line into a parallel tangent of the big circle.

Step 1 — in directed angles $\angle(AQ, AB) = \angle(AQ, AP)$, because $AB$ and $AP$ are the same line, and $\angle(AQ, AP)$ is the inscribed angle of $\omega_1$ on the chord $QP$: it does not depend on where $A$ is. Likewise $\angle(BQ, BA) = \angle(BQ, BP)$ is the inscribed angle of $\omega_2$ on $QP$. Two angles of $QAB$ are fixed, so all these triangles are directly similar.

Step 2 — in complex numbers, direct similarity of $QAB$ to a fixed triangle means $b - q = \lambda(a - q)$ for one fixed complex $\lambda$: the map $A \mapsto B$ is the spiral similarity about $Q$ with multiplier $\lambda$, and it carries $\omega_1$ onto $\omega_2$.

Step 3 — the midpoint is $n = \tfrac{a + b}{2} = q + \tfrac{1 + \lambda}{2}(a - q)$. So $A \mapsto N$ is again a spiral similarity about $Q$, with multiplier $\tfrac{1 + \lambda}{2}$, which is not $0$: $\lambda = -1$ would put $Q$ on every line $AB$, but those lines pass through $P$. A spiral similarity sends the circle $\omega_1$ to a circle, and that circle is the locus. $\blacksquare$

How to find it. One fixed point ($Q$) and a line that moves through another ($P$): the triangle on the fixed point never changes shape. Whenever a problem says “a variable line through $P$”, look for the spiral similarity at the other intersection.

Step 1 — let $M \ne F$ be the second intersection of the circles $(FAD)$ and $(FBC)$. Chasing directed angles,

using the circle $(FAD)$ for the first term and $(FBC)$ for the second.

Step 2 — $D$, $F$, $C$ are collinear, so $DF$ and $CF$ are the same line $CD$, and the sum telescopes: $\angle(DA, CD) + \angle(CD, CB) = \angle(DA, CB)$. Line $DA$ passes through $E$ and so does $CB$, so this is $\angle(EA, EB)$. Hence $\angle(MA, MB) = \angle(EA, EB)$ and $M$ lies on $(EAB)$. The identical chase with $C, D$ in place of $A, B$ puts $M$ on $(ECD)$.

Step 3 — by the spiral-similarity lemma applied to $A \mapsto D$, $B \mapsto C$ (lines $AD$ and $BC$ meet at $E$), the center is the second intersection of $(EAB)$ and $(EDC)$, which is $M$. $\blacksquare$

How to find it. Four lines give four triangles; circles through three points each are begging for a common point. Define $M$ from two circles and prove it is on the others — never try to prove four circles concur “at once”.