#2 Power of a Point

By the two-secant theorem, \(5\cdot18=6\cdot PD\). Hence \(PD=15\).

We have \(PT^2=3\cdot27=81\). Therefore \(PT=9\).

By the intersecting-chords theorem, \(XA\cdot XB=XC\cdot XD\). Hence \(8\cdot6=4\cdot XD\), so \(XD=12\).

By the tangent formula, the power of point \(P\) with respect to the circle equals \(PA^2\). The same power also equals \(PB^2\). Therefore \(PA^2=PB^2\), and since lengths are positive, \(PA=PB\).

Join \(A\) to \(C\), and \(B\) to \(D\). Consider triangles \(PAC\) and \(PDB\). The angle at \(P\) is common, because \(P,A,B\) and \(P,C,D\) are collinear. Also, \(\angle PAC=\angle BAC\), \(\angle PDB=\angle CDB\), and angles \(\angle BAC\) and \(\angle CDB\) are equal as inscribed angles standing on chord \(BC\). Therefore \(\triangle PAC\sim\triangle PDB\). From similarity, \(\frac{PA}{PD}=\frac{PC}{PB}\), hence \(PA\cdot PB=PC\cdot PD\).

Let the circle through \(A,B,C\) meet ray \(PC\) for the second time at \(D'\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). By the condition, \(PA\cdot PB=PC\cdot PD\), hence \(PD'=PD\). Since \(D\) and \(D'\) lie on the same ray from \(P\), they coincide. Therefore \(A,B,C,D\) are cyclic.

Let \(XA=2k\), \(XB=5k\). Then \(XA\cdot XB=10k^2\). By the chord theorem, \(10k^2=XC\cdot XD=6\cdot15=90\). Hence \(k^2=9\), \(k=3\). Therefore \(XA=6\), \(XB=15\).

Let \(PA=x\). Then \(PB=x+20\). By the tangent-secant formula, \(12^2=x(x+20)\), so \(x^2+20x-144=0\). The positive root is \(x=6\). Hence \(PA=6\).

Lines \(AB\) and \(AC\) are secants of circle \((BCDE)\). Therefore \(AD\cdot AB=AE\cdot AC\). We get \(4\cdot14=7\cdot AC\), hence \(AC=8\).

For the first circle, the power of point \(P\) equals \(PA\cdot PB\). For the second circle, the intersection points with line \(PAB\) are the same \(A\) and \(B\), so the power also equals \(PA\cdot PB\). Hence the powers are equal.

By the tangent-secant formula, \(PT^2=PA\cdot PB\). Thus \(225=9\cdot PB\), so \(PB=25\). The secant passes through the centre, hence \(AB\) is a diameter. We get \(AB=PB-PA=25-9=16\).

By the secant theorem, \(PA\cdot PB=PC\cdot PD\). Hence \(\frac{PD}{PB}=\frac{PA}{PC}=\frac{2}{3}\). Therefore \(PD:PB=2:3\).

From point \(A\), two secants to circle \((BCDE)\) are drawn: \(ADB\) and \(AEC\). Therefore \(AD\cdot AB=AE\cdot AC\). Dividing both sides by \(AE\cdot AB\), we get \(\frac{AD}{AE}=\frac{AC}{AB}\).

On ray \(AB\) lie points \(D,B\), and on ray \(AC\) lie points \(E,C\). The equality \(AD\cdot AB=AE\cdot AC\) has the form of equality of products of two secants from point \(A\). By the converse power criterion, points \(D,B,E,C\), that is \(B,C,D,E\), lie on one circle.

Diagonals \(AC\) and \(BD\) are chords of one circle and meet at \(P\). By the intersecting-chords theorem, \(PA\cdot PC=PB\cdot PD\).

Line \(PAB\) meets the circle at \(A\) and \(B\), while line \(PCD\) meets it at \(C\) and \(D\). Therefore, by the two-secant theorem, \(PA\cdot PB=PC\cdot PD\).

By the tangent-secant theorem, \(TA^2=TB\cdot TC\). By the tangent-chord theorem, \(\angle TAB=\angle ACB\) and \(\angle TAC=\angle ABC\), so \(\triangle TAB\sim\triangle TCA\). From similarity, \(\frac{TB}{TA}=\frac{AB}{AC}\) and \(\frac{TA}{TC}=\frac{AB}{AC}\). Multiplying, we get \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\).

With respect to circle \((ABD)\), from point \(C\) we have tangent \(CE\) and secant \(CDB\). Since \(D\) lies on side \(BC\), the points on the secant are in the order \(C,D,B\). By the tangent-secant formula, \(CE^2=CD\cdot CB\).

The power of point \(P\) with respect to the first circle equals \(PT_1^2\), and with respect to the second equals \(PT_2^2\). If \(PT_1=PT_2\), then \(PT_1^2=PT_2^2\). Therefore the powers are equal.

Draw the circle through \(A,B,C\). Let it meet line \(l_2\) for the second time at \(D'\), located on the same ray from \(P\) as \(D\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). Comparing with the condition, we get \(PD'=PD\), hence \(D'=D\). Therefore \(A,B,C,D\) lie on one circle.

By the result proved above for a tangent to the circumcircle, \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\). For completeness: it follows from the similarity \(\triangle TAB\sim\triangle TCA\), obtained by the tangent-chord theorem. For \(AB=9\), \(AC=6\), we get \(TB:TC=9^2:6^2=81:36=9:4\).

Equality of powers for point \(X\) gives \(XO_1^2-r_1^2=XO_2^2-r_2^2\), that is \(XO_1^2-XO_2^2=r_1^2-r_2^2\). The same equality holds for point \(Y\). Thus points \(X\) and \(Y\) have the same difference of squares of distances to \(O_1\) and \(O_2\). The locus of such points is a line perpendicular to \(O_1O_2\). Therefore \(XY\perp O_1O_2\).

Assume that \(PT\) is not tangent. Then line \(PT\) meets the circle at another point \(U\). By the secant theorem, \(PT\cdot PU=PA\cdot PB\). By the condition, \(PT^2=PA\cdot PB\), hence \(PT\cdot PU=PT^2\), so \(PU=PT\). But on the ray from \(P\), the point of the circle at distance \(PT\) is already \(T\), so \(U=T\), contradicting two distinct intersections. Therefore \(PT\) is tangent to the circle.

For the first circle, the power of point \(P\) equals \(PC\cdot PD\). For the second circle, it equals \(PE\cdot PF\). But \(P\) lies on line \(AB\), and \(A\) and \(B\) are common points of the circles. Therefore both powers are also equal to the same product \(PA\cdot PB\). Hence \(PC\cdot PD=PE\cdot PF\).