#3 Radical Axis

For any point \(X\) on line \(AB\), secant \(XAB\) meets both circles at the same points \(A\) and \(B\). Therefore the power of \(X\) with respect to each circle equals \(XA\cdot XB\). Hence the powers are equal, and \(AB\) is the radical axis.

Let \(X\) lie on the common tangent at \(A\). With respect to the first circle, \(XA\) is a tangent, so the power equals \(XA^2\). With respect to the second circle, \(XA\) is also a tangent, so the power is again \(XA^2\). Hence the powers are equal for all points of this line.

Let the common radius be \(r\). Equality of powers has the form \(XO_1^2-r^2=XO_2^2-r^2\), hence \(XO_1^2=XO_2^2\), so \(XO_1=XO_2\). The locus of points equidistant from \(O_1\) and \(O_2\) is the perpendicular bisector of \(O_1O_2\).

Let \(O_1H=x\). Then \(O_2H=13-x\). Equality of powers gives \(x^2-5^2=(13-x)^2-8^2\). We get \(x^2-25=x^2-26x+169-64\), hence \(-25=105-26x\), so \(x=5\). Therefore \(O_1H=5\).

The power of point \(P\) with respect to \(\omega_1\) equals \(PT_1^2\), and with respect to \(\omega_2\) equals \(PT_2^2\). Since \(PT_1=PT_2\), the powers are equal. Thus \(P\) lies on the radical axis of the circles. They intersect at \(A\) and \(B\), so the radical axis is line \(AB\). Therefore \(P,A,B\) are collinear.

The product \(PA\cdot PB\) is the power of point \(P\) with respect to \(\omega_1\). The product \(PC\cdot PD\) is the power of point \(P\) with respect to \(\omega_2\). By the condition these powers are equal, so \(P\) lies on the radical axis of the two circles.

From \(R\in\operatorname{Rad}(\omega_1,\omega_2)\), we get \(\operatorname{Pow}_{\omega_1}(R)=\operatorname{Pow}_{\omega_2}(R)\). From \(R\in\operatorname{Rad}(\omega_2,\omega_3)\), we get \(\operatorname{Pow}_{\omega_2}(R)=\operatorname{Pow}_{\omega_3}(R)\). Therefore \(\operatorname{Pow}_{\omega_1}(R)=\operatorname{Pow}_{\omega_3}(R)\), so \(R\) lies on the third radical axis.

The first common chord is the radical axis of the first and second circles, and the second is the radical axis of the second and third. Their intersection point \(R\) has equal powers with respect to the first, second, and third circles. Therefore \(R\) lies on the radical axis of the first and third circles. Since they intersect, this axis is their common chord.

For point \(P\), equality of tangent lengths means equality of powers with respect to the two circles, so \(P\) lies on their radical axis. Similarly, \(Q\) lies on the same radical axis. Since the circles intersect at \(A\) and \(B\), the radical axis is line \(AB\). Therefore \(P,Q,A,B\) are collinear.

Since \(P\) lies on the radical axis, \(\operatorname{Pow}_{\omega_1}(P)=\operatorname{Pow}_{\omega_2}(P)\). But \(\operatorname{Pow}_{\omega_1}(P)=PA\cdot PB\), and \(\operatorname{Pow}_{\omega_2}(P)=PC\cdot PD\). Therefore \(PA\cdot PB=PC\cdot PD\).

Equality of powers at point \(H\) gives \(x^2-13^2=(20-x)^2-7^2\). Then \(x^2-169=x^2-40x+400-49\), hence \(-169=351-40x\), so \(40x=520\) and \(x=13\). Therefore \(O_1H=13\).

Let the centres be \(O_1,O_2\) and the radii \(r_1,r_2\). For a point \(X\) on the radical axis, \(XO_1^2-r_1^2=XO_2^2-r_2^2\), that is \(XO_1^2-XO_2^2=r_1^2-r_2^2\). The locus of points with constant difference of squares of distances to \(O_1\) and \(O_2\) is a line perpendicular to \(O_1O_2\). Hence the radical axis is perpendicular to the line of centres.

Line \(AB\) is the radical axis of \(\omega_1\) and \(\omega_2\), and \(CD\) is the radical axis of \(\omega_2\) and \(\omega_3\). Their intersection \(R\) is the radical center of the three circles. Therefore \(R\) lies on the radical axis of \(\omega_1\) and \(\omega_3\). Since these circles intersect at \(E,F\), their radical axis is line \(EF\). Hence \(R,E,F\) are collinear.

The set of points having equal powers with respect to two given circles is, by definition, their radical axis. The radical axis is a line. Therefore all points \(X,Y,Z\) lie on this line.

From orthogonality of \(\gamma\) and \(\omega_1\), we get \(XO_1^2= ho^2+r_1^2\), hence \(XO_1^2-r_1^2= ho^2\). Similarly, \(XO_2^2-r_2^2= ho^2\). Therefore the powers of point \(X\) with respect to \(\omega_1\) and \(\omega_2\) are equal, and \(X\) lies on the radical axis.

Let the centres of \(\gamma_1\) and \(\gamma_2\) be \(X_1\) and \(X_2\). Since \(\gamma_1\) is orthogonal to both given circles, by the previous fact \(X_1\) lies on their radical axis. Similarly, \(X_2\) lies on the same radical axis. Therefore both centres lie on one radical axis.

Since \(P\) lies on the radical axis, the powers of point \(P\) with respect to the two circles are equal. But these powers are \(PT_1^2\) and \(PT_2^2\). Hence \(PT_1^2=PT_2^2\), and since lengths are positive, \(PT_1=PT_2\).

Equality of tangent lengths from point \(P\) means equality of powers of \(P\) with respect to the two circles, so \(P\) lies on the radical axis. Similarly, \(Q\) lies on the same axis. Therefore line \(PQ\) is the radical axis. The radical axis of two nonconcentric circles is perpendicular to the line of centres, hence \(PQ\perp O_1O_2\).

The product \(RA_i\cdot RB_i\) equals the power of point \(R\) with respect to circle \(\omega_i\). By the condition, all three powers are equal. Hence \(R\) has equal powers with respect to \(\omega_1\) and \(\omega_2\), with respect to \(\omega_2\) and \(\omega_3\), and with respect to \(\omega_1\) and \(\omega_3\). Therefore \(R\) lies on all three radical axes.

The common tangent at \(A\) is the radical axis of \(\omega_1\) and \(\omega_2\), so \(R\) lies on this axis. The common tangent at \(B\) is the radical axis of \(\omega_2\) and \(\omega_3\), so \(R\) lies on it as well. By the radical center theorem, \(R\) lies on the radical axis of \(\omega_1\) and \(\omega_3\).

Since \(R\) is the radical center, its powers with respect to \(\omega_1,\omega_2,\omega_3\) are equal. For each circle, the power of point \(R\) equals the product of the segments of any secant through \(R\): \(RA_i\cdot RB_i\). Therefore all three products are equal.

Since the common power equals \(\rho^2\), we have \(XO_1^2-r_1^2= ho^2\) and \(XO_2^2-r_2^2= ho^2\). Hence \(XO_1^2= ho^2+r_1^2\) and \(XO_2^2= ho^2+r_2^2\). This is exactly the condition that the circle with centre \(X\) and radius \(\rho\) is orthogonal to each of the given circles.

Let \(X\) be the centre of one of the circles orthogonal to \(\omega_1\) and \(\omega_2\). By the problem about the centre of an orthogonal circle, \(X\) lies on the radical axis of \(\omega_1\) and \(\omega_2\). The same is true for the centre of any other such circle. Therefore all these centres lie on one line: the radical axis of the fixed circles.

Let the radius of \(\gamma\) be \(\rho\). If \(\gamma\) is orthogonal to \(\omega_i(O_i,r_i)\), then \(XO_i^2= ho^2+r_i^2\), hence \(XO_i^2-r_i^2= ho^2\). Thus the powers of point \(X\) with respect to all three circles are equal. Therefore \(X\) lies on every pairwise radical axis, so it is the radical center.