With the definition, we can derive trivial relationship between the power of a point and the location of the point. If \(P_\omega (P) < 0\), then \(P\) is inside \(\omega \). If \(P_\omega (P) = 0\), then \(P\) is on \(\omega \) and \(P\) is outside \(\omega \) if and only if \(P_\omega (P) > 0\).

Proof.

Let \(\omega _1\) be a circle defined as \((x - a_1)^2 + (y - b_1)^2 = r_1^2\). Similarly, let \(\omega _2\) be a circle \((x - a_2)^2 + (y - b_2)^2 = r_1^2\). Let \(P(x, y)\) be any points such that \(P_{\omega _1}(P) = P_{\omega _2}(P)\). Substituting, the following equation holds by definition. \[ (x - a_1)^2 + (y - b_1)^2 - r_1^2 = (x - a_2)^2 + (y - b_2)^2 - r_2^2 \] Expanding and rearranging the equation, \begin{align*} &\quad \ x^2 - 2a_1 x + a_1^2 + y^2 - 2b_1 y + b_1^2 - r_1^2 \\ &= x^2 - 2a_2 x + a_2^2 + y^2 - 2b_2 y + b_2^2 - r_2^2 \\ &(2a_2 - 2a_1) x + (2b_2 - 2b_1) y + C = 0 \end{align*}

holds where \(C\) is the constant values. Notice that the locus of all \((x, y)\) that satisfies the equation above is a linear function. Thus, the radical axis is always linear.

Proof.

Continuing with the equation \[ (2a_2 - 2a_1) x + (2b_2 - 2b_1) y + C = 0 \] from the previous proof, it is evident that the slope of the line is \(-\frac {2a_2 - 2a_1}{2b_2 - 2b_1}\). Continuing, the slope of the line that passes through the two centers of \(\omega _1\) and \(\omega _2\) is \(\frac {b_2 - b_1}{a_2 - a_1}\). Because \(-\frac {2a_2 - 2a_1}{2b_2 - 2b_1} \cdot \frac {b_2 - b_1}{a_2 - a_1} = -1\), the two lines are perpendicular to each other.

Proof.

Let the centers of \(\omega _1, \omega _2, \omega _3\) be \(O_1, O_2, O_3\) and the radical axis be \(l\). By definition, the radical axis of \(\omega _1\) and \(\omega _2\) is \(l\) and \(O_1O_2 \perp l\). Similarly, \(O_1O_3 \perp l\) since \(l\) is the radical axis of \(\omega _1\) and \(\omega _3\). Because there exists exactly one line perpendicular to \(l\) that passes through \(O_1\), \(O_1O_2\) and \(O_1O_3\) lines are the same line and \(O_1,O_2,O_3\) are collinear.

Never underestimate! Facts like \(OA = OB\) for a circle with center \(O\) and \(P_\omega (A) = P_\omega (B)\) can be the key to the solution. Whenever you see three circles, circles defined by reflection, collinearity of their centers, and/or the intersection of three circles, we should be reminded of coaxial circles and suspect it as the key at least once.

Proof.

Let \(\omega _1, \omega _2, \omega _3\) be three distinct circles with the centers \(O_1, O_2, O_3\). Let \(l_1, l_2, l_3\) be the radical axes of the pairs \(O_1, O_2\) and \(O_2, O_3\) and \(O_3, O_1\) respectively. Let \(K\) be the intersection of \(l_1\) and \(l_2\). By definition, \(P_{\omega _1}(K) = P_{\omega _2}(K)\) and \(P_{\omega _2}(K) = P_{\omega _3}(K)\). Therefore, \(P_{\omega _1}(K) = P_{\omega _3}(K)\) and \(K \in l_3\). Therefore, \(l_1, l_2, l_3\) are concurrent.

Proof.

First, notice that for any circles \(x^2 + y^2 + ax + by + c = 0\) and a point \(A(x_1, y_1)\), \(P_\omega (A) = (x_1 - p)^2 + (y_2 - q)^2 - r^2 = x_1^2 + y_1^2 + ax_1 + by_1 + c\). Moreover, if \(\frac {P_{\omega _1}(A)}{P_{\omega _2}(A)} = \frac {P_{\omega _1}(B)}{P_{\omega _2}(B)} = \frac {P_{\omega _1}(C)}{P_{\omega _2}(C)} = k\), then \(P_{\omega _1}(X) - k P_{\omega _2}(X) = 0\) for \(X = A, B, C\).

Let \(\omega _1\) be defined as \(x^2 + y^2 + a_1x + b_1y + c_1 = 0\) and \(\omega _2\) as \(x^2 + y^2 + a_2x + b_2y + c_2 = 0\). Rewriting the expression \(P_{\omega _1}(X) - k P_{\omega _2}(X)\), the following equations are obtained. \begin{align*} P_{\omega _1}(X) - k P_{\omega _2}(X) &= x^2 + y^2 + a_1x + b_1y + c_1 \\ &\qquad \qquad \qquad \,- k(x^2 + y^2 + a_2x + b_2y + c_2) \\ &= (1 - k) x^2 + (1 - k) y^2 + (a_1 - ka_2) x \\ &\qquad \qquad \qquad \,+ (b_1 - kb_2) y + (c_1 - kc_2) \end{align*}

Define \(\omega _3\) as \((1 - k) x^2 + (1 - k) y^2 + (a_1 - ka_2) x + (b_1 - kb_2) y + (c_1 - kc_2) = 0\) for \(k \neq 1\). Then, \(A, B, C \in \omega _3\) and \(\omega _3\) is the circumcircle of \(\triangle {ABC}\). Moreover by construction, the power of any point \(X\) satisfy the following equation. \[ P_{\omega _3}(X) = \frac {P_{\omega _1}(X) - k P_{\omega _2}(X)}{1 - k} \] Therefore, if \(X\) lies on the radical axis of \(\omega _3\) and \(\omega _1\), i.e. \(P_{\omega _3}(X) = P_{\omega _1}(X)\), then \((1 - k) P_{\omega _1}(X) = P_{\omega _1}(X) - k P_{\omega _2}(X)\) and \(P_{\omega _1}(X) = P_{\omega _2}(X)\). In other words, \(P_{\omega _3}(X) = P_{\omega _1}(X)\) if and only if \(P_{\omega _1}(X) = P_{\omega _2}(X)\) and \(\omega _1, \omega _2, \omega _3\) are coaxial circles.

If \(k = 1\), then \(P_{\omega _1}(A) = P_{\omega _2}(A)\), \(P_{\omega _1}(B) = P_{\omega _2}(B)\), and \(P_{\omega _1}(C) = P_{\omega _2}(C)\) holds and \(A, B, C\) lies on the radical axis of \(\omega _1\) and \(\omega _2\). Because radical axis is a straight line, \(A, B, C\) are collinear and the points form the radical axis.

Solution.

First, let’s draw the diagram.

Notice that because \(Q\) and \(R\) both lies of the circles \(\omega _1\) (left) and \(\omega _2\) (right), it suffices to show that \(D\) lies on the radical axis of \(\omega _1\) and \(\omega _2\). In other words, we must show that \(P_{\omega _1} (D) = P_{\omega _2} (D)\). Consider the following equation. \[ P_{\omega _1} (D) = \overline {AD} \cdot \overline {PD} = \overline {BD} \cdot (2 \cdot \overline {MD}) = 2 \overline {BD} \cdot \overline {MD} = \overline {DC} \cdot \overline {MD} \] Therefore, it suffices to show that \(P_{\omega _2} (D) = \overline {DC} \cdot \overline {MD}\).

To find \(P_{\omega _2}\), we can use another classic pattern that we can see from the problem. We see the point \(D\), the incenter \(J\) and the circumcircle of \(\triangle {MJC}\). Naturally trying to implement incenter/excenter lemma here, we can draw and find excenter of \(\triangle {MDC}\) at \(D\).

Notice that \(\angle {MDJ} = \angle {JDC}\) by construction. Similarly, \(\angle {MI_DJ} = \angle {MCJ} = \angle {JCD}\). Thus \(\triangle {MDI_D}\) and \(\triangle {JDC}\) are similar. Because \(MD : JD = DI_D : DC\), \(P_{\omega _2}(D) = DJ \cdot DI_D = MD \cdot DC\).