Circles: Chords and Tangents: Very Difficult Problems with Solutions

Solution:
In [tex]k_1[/tex] every chord through [tex]P[/tex] gives the same product of segments, and the chord [tex]AB[/tex] gives [tex]PA \times PB[/tex]. In [tex]k_2[/tex] both chords [tex]AB[/tex] and [tex]CD[/tex] pass through [tex]P[/tex], so [tex]PA \times PB[/tex] = [tex]PC \times PD[/tex]. In [tex]k_3[/tex] the chord [tex]CD[/tex] gives [tex]PC \times PD[/tex].
Let the line [tex]PE[/tex] meet [tex]k_1[/tex] again at [tex]F_1[/tex] and [tex]k_3[/tex] again at [tex]F_3[/tex]. The products in [tex]k_1[/tex] and [tex]k_3[/tex] are equal, so [tex]PE \times PF_1=PE \times PF_3[/tex], hence [tex]PF_1=PF_3[/tex] and [tex]F_1[/tex] = [tex]F_3[/tex]. This point is common to [tex]k_1[/tex] and [tex]k_3[/tex] and differs from [tex]E[/tex], so it is [tex]F[/tex].
Therefore the line [tex]EF[/tex] passes through [tex]P[/tex].

Solution:
Extend [tex]CL[/tex] to meet the circumscribed circle again at [tex]M[/tex]. Then [tex]\angle ACL=\angle MCB[/tex] ([tex]CL[/tex] is the bisector) and [tex]\angle CAL=\angle CMB[/tex] (inscribed angles on the same arc), so [tex]\triangle ACL \sim \triangle MCB[/tex].
Hence [tex]\frac{AC}{CM}=\frac{CL}{CB}[/tex], i.e. [tex]AC \times CB=CL \times CM=CL(CL+LM)=CL^2+CL \times LM[/tex].
By the intersecting chords theorem [tex]CL \times LM=LA \times LB[/tex], so [tex]CL^2=AC \times CB-LA \times LB[/tex].

Solution:
The trapezoid is isosceles; the projection of a leg is [tex]\frac{15-1}{2}=7[/tex], so [tex]h=\sqrt{14^2-7^2}=7\sqrt{3}[/tex]. Place [tex]A(0,0),\ B(15,0),\ C(8,7\sqrt{3}),\ D(7,7\sqrt{3})[/tex].
The center of the circle is the midpoint of [tex]BC[/tex]: [tex]Q\left(\frac{23}{2},\frac{7\sqrt{3}}{2}\right)[/tex], and the radius is [tex]7[/tex]. The line is [tex]AD:\ y=\sqrt{3}x[/tex], so the distance from [tex]Q[/tex] to it is [tex]d=\frac{\left|\sqrt{3} \times \frac{23}{2}-\frac{7\sqrt{3}}{2}\right|}{2}=4\sqrt{3}[/tex].
[tex]MN=2\sqrt{7^2-(4\sqrt{3})^2}=2\sqrt{49-48}=2[/tex].

Solution:
Let [tex]M[/tex] and [tex]N[/tex] be the projections of [tex]O_1[/tex] and [tex]O_2[/tex] on the line. The perpendicular from the center bisects the chord, so [tex]AM=MP;\ AN=NQ[/tex], hence [tex]PQ=2MN[/tex].
[tex]MN[/tex] is the projection of [tex]O_1O_2[/tex] on the line, so [tex]MN \le O_1O_2[/tex], with equality exactly when the line is parallel to [tex]O_1O_2[/tex].
Therefore [tex]PQ[/tex] is largest, equal to twice [tex]O_1O_2[/tex], when the line is parallel to [tex]O_1O_2[/tex].

Solution:
Draw the common tangent [tex]t[/tex] at [tex]T[/tex]. The angle between a tangent and a chord equals the inscribed angle on the same arc, so in [tex]k_1[/tex]: [tex]\angle TAB=\angle(t;TB)[/tex].
[tex]\angle(t;TB)=\angle(t;TD)[/tex] (vertical angles, since [tex]B[/tex], [tex]T[/tex], [tex]D[/tex] are collinear), and in [tex]k_2[/tex]: [tex]\angle(t;TD)=\angle TCD[/tex].
So [tex]\angle TAB=\angle TCD[/tex]; these are alternate angles at the transversal [tex]AC[/tex], hence [tex]AB \parallel CD[/tex].

Solution:
For two externally tangent circles touching the same line, the distance between the points of tangency is twice the geometric mean of the radii: [tex]T_1T_2=2\sqrt{R_1R_2}[/tex], and similarly [tex]T_1T=2\sqrt{R_1r};\ TT_2=2\sqrt{rR_2}[/tex].
Since [tex]T[/tex] lies between [tex]T_1[/tex] and [tex]T_2[/tex], [tex]2\sqrt{R_1r}+2\sqrt{rR_2}=2\sqrt{R_1R_2}[/tex]; dividing by the product of the three square roots, [tex]\frac{1}{\sqrt{r}}=\frac{1}{\sqrt{R_1}}+\frac{1}{\sqrt{R_2}}[/tex].
Hence [tex]r=\frac{R_1R_2}{\left(\sqrt{R_1}+\sqrt{R_2}\right)^2}[/tex].

Solution:
Place [tex]A(0,0),\ B(R-r,0)[/tex] (the circles touch at a point of the positive x-axis), and let the center be [tex]Q(x,\rho)[/tex]. The tangency conditions give [tex]AQ=R-\rho[/tex] and [tex]BQ=r+\rho[/tex].
From the first, [tex]x^2=R^2-2R\rho[/tex]. From the second, [tex](x-R+r)^2+\rho^2=(r+\rho)^2[/tex]; subtracting and using [tex]x^2=R^2-2R\rho[/tex]: [tex]x=R-\frac{\rho(R+r)}{R-r}[/tex].
Substituting into [tex]x^2=R^2-2R\rho[/tex]: [tex]\frac{\rho(R+r)^2}{(R-r)^2}=\frac{4Rr}{R-r}[/tex], so [tex]\rho=\frac{4Rr(R-r)}{(R+r)^2}[/tex].

Solution:
The homothety with center [tex]M[/tex] and ratio [tex]\frac{R_1}{R_2}[/tex] ([tex]R_1[/tex] and [tex]R_2[/tex] are the radii of [tex]k_1[/tex] and [tex]k_2[/tex]) maps [tex]k_2[/tex] onto [tex]k_1[/tex]. It maps [tex]L[/tex] to the second intersection point [tex]N[/tex] of the line [tex]ML[/tex] with [tex]k_1[/tex], and the tangent [tex]AB[/tex] to [tex]k_2[/tex] at [tex]L[/tex] to the tangent [tex]t[/tex] to [tex]k_1[/tex] at [tex]N[/tex], so [tex]t \parallel AB[/tex].
A tangent parallel to a chord touches the circle at the midpoint of the arc cut off by the chord, so [tex]\overset{\frown}{AN}=\overset{\frown}{NB}[/tex].
Equal arcs subtend equal inscribed angles: [tex]\angle AMN=\angle NMB[/tex], i.e. [tex]ML[/tex] bisects [tex]\angle AMB[/tex].

Solution:
[tex]O_1[/tex] lies on the bisector of the angle at [tex]C[/tex], which passes through [tex]O[/tex], and [tex]\angle O_1CA=30^\circ[/tex]; so the radius (the distance from [tex]O_1[/tex] to the sides) is half of [tex]CO_1[/tex]: [tex]CO_1=2\rho[/tex].
The circle touches [tex]k(O;R)[/tex] internally, so [tex]OO_1=R-\rho[/tex]; [tex]O_1[/tex] lies on [tex]CO[/tex] beyond [tex]O[/tex], so [tex]CO_1=CO+OO_1[/tex], i.e. [tex]2\rho=R+(R-\rho)[/tex].
Hence [tex]\rho=\frac{2R}{3}[/tex].

Solution:
Use the curvatures (reciprocals of the radii): [tex]k_1=\frac{1}{a};\ k_2=\frac{1}{b};\ k_3=\frac{1}{c};\ k_4=\frac{1}{r}[/tex]. By Descartes' circle theorem, for four mutually tangent circles [tex]k_4=k_1+k_2+k_3+2\sqrt{k_1k_2+k_2k_3+k_3k_1}[/tex] (plus sign for the small inner circle).
Here [tex]k_1k_2+k_2k_3+k_3k_1=\frac{a+b+c}{abc}[/tex], so [tex]\frac{1}{r}=\frac{ab+bc+ca+2\sqrt{abc(a+b+c)}}{abc}[/tex], and [tex]r=\frac{abc}{ab+bc+ca+2\sqrt{abc(a+b+c)}}[/tex].

Solution:
Since [tex]\angle ADC=90^\circ[/tex], [tex]D[/tex] lies on the circle with diameter [tex]AC[/tex], so [tex]MD=MA=MC=7[/tex], and the isosceles triangle gives [tex]\angle MDC=\angle MCD=\angle C[/tex].
With the altitude [tex]BH[/tex] of [tex]\triangle ABC[/tex]: [tex]BH=\sqrt{25^2-7^2}=24[/tex], so [tex]\sin C=\frac{BH}{BC}=\frac{24}{25}[/tex].
The angle between the tangent [tex]BC[/tex] and the chord [tex]MD[/tex] equals the inscribed angle on [tex]MD[/tex], so by the Law of Sines [tex]r=\frac{MD}{2\sin C}=\frac{7}{2 \times \frac{24}{25}}=\frac{175}{48}[/tex].

Solution:
Place [tex]O_1(0,0),\ O_2(2\sqrt{2},0)[/tex]. The tangent [tex]l[/tex] is [tex]l:\ \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y=3[/tex] (its distances to the centers equal the radii), and the tangent segment is [tex]AB=\sqrt{(2\sqrt{2})^2-(3-1)^2}=2[/tex].
Solving [tex]x^2+y^2=9;\ (x-2\sqrt{2})^2+y^2=1[/tex] gives the intersection points; the one nearer to [tex]l[/tex] is [tex]C(2\sqrt{2},1)[/tex], at distance [tex]d=3-\frac{\sqrt{2}}{2}(2\sqrt{2}+1)=1-\frac{\sqrt{2}}{2}[/tex] from [tex]l[/tex].
So [tex]S=\frac{AB \times d}{2}=1-\frac{\sqrt{2}}{2}=\frac{2-\sqrt{2}}{2}[/tex].

Solution:
Let [tex]O[/tex] be the center, [tex]OM=m[/tex], and [tex]d[/tex] the distance from [tex]O[/tex] to the chord: [tex]d=m\sin 45^\circ=\frac{m}{\sqrt{2}}[/tex]. Then [tex]CD^2=4(R^2-d^2)=4R^2-2m^2[/tex].
By the intersecting chords theorem (with the diameter through [tex]M[/tex]) [tex]CM \times DM=R^2-m^2[/tex].
Hence [tex]CM^2+DM^2=(CM+DM)^2-2CM \times DM=4R^2-2m^2-2(R^2-m^2)=2R^2[/tex], which does not depend on the position of [tex]M[/tex].

Solution:
By the angle bisector theorem [tex]\frac{BL}{LC}=\frac{AB}{AC}=\frac{5}{7}[/tex], so [tex]BL=\frac{35}{4};\ LC=\frac{49}{4}[/tex].
By the formula for the bisector proved earlier on this page, [tex]AL^2=AB \times AC-BL \times LC=560-\frac{1715}{16}=\frac{7245}{16}[/tex]. By the intersecting chords theorem [tex]AL \times LL'=BL \times LC=\frac{1715}{16}[/tex].
Hence [tex]\frac{AL}{LL'}=\frac{AL^2}{AL \times LL'}=\frac{7245}{1715}=\frac{207}{49}[/tex].

Solution:
The lines [tex]AD;\ BC[/tex] meet at [tex]P[/tex], and [tex]\frac{PD}{PA}=\frac{PC}{PB}=\frac{CD}{AB}=\frac{2}{3}[/tex], so [tex]PA=15;\ PB=36[/tex] and [tex]PD=10[/tex]. Since [tex]PA^2+PB^2=225+1296=1521=AB^2[/tex], [tex]\angle APB=90^\circ[/tex].
Let [tex]T[/tex] be the point of tangency on [tex]BC[/tex]; by the tangent–secant theorem [tex]PT^2=PD \times PA=10 \times 15=150[/tex].
Place [tex]P(0,0),\ A(15,0),\ T(0,5\sqrt{6})[/tex]; the center is [tex]Q(\rho,5\sqrt{6})[/tex], and [tex](15-\rho)^2+150=\rho^2[/tex] gives [tex]\rho=\frac{25}{2}[/tex].

Solution:
Let [tex]AB=x[/tex], so [tex]AD=x+24[/tex]. By the Law of Cosines in [tex]\triangle ABD[/tex]: [tex]56^2=x^2+(x+24)^2-x(x+24)[/tex], i.e. [tex]x^2+24x-2560=0[/tex], so [tex]x=40[/tex] and [tex]AD=64[/tex].
The angle between the tangent and the chord [tex]BC[/tex] equals the inscribed angle on it: [tex]\angle ABC=\angle ADB[/tex]; with the common angle at [tex]A[/tex], [tex]\triangle ABC \sim \triangle ADB[/tex].
So [tex]\frac{BC}{DB}=\frac{AB}{AD}[/tex], and [tex]BC=\frac{56 \times 40}{64}=35[/tex].

Solution:
By the tangent–secant theorem from [tex]B[/tex]: [tex]BA^2=BM \times BC=\frac{b}{2} \times b[/tex], so [tex]AB=\frac{b}{\sqrt{2}}[/tex]. Let [tex]\alpha[/tex] be the base angle: [tex]\cos\alpha=\frac{AB}{2b}=\frac{1}{2\sqrt{2}}[/tex], [tex]\sin\alpha=\sqrt{1-\frac{1}{8}}=\frac{\sqrt{14}}{4}[/tex].
The angle between the tangent [tex]AB[/tex] and the chord [tex]AC[/tex] equals the inscribed angle on it: [tex]\angle AMC=\angle BAC=\alpha[/tex]. By the Law of Sines in this circle [tex]R=\frac{AC}{2\sin\alpha}=\frac{2b}{\sqrt{14}}[/tex], so [tex]b=\frac{R\sqrt{14}}{2}[/tex].

Solution:
The external bisector [tex]CL'[/tex] at [tex]C[/tex] meets the line [tex]AB[/tex] at [tex]L'[/tex] beyond [tex]B[/tex], and [tex]\frac{AL'}{BL'}=\frac{AC}{BC}=\frac{AL}{BL}=\frac{5}{3}[/tex], so [tex]\frac{8+BL'}{BL'}=\frac{5}{3}[/tex], i.e. [tex]BL'=12[/tex].
The internal and external bisectors are perpendicular: [tex]\angle LCL'=90^\circ[/tex], so [tex]C[/tex] lies on the circle with diameter [tex]LL'=3+12=15[/tex], whose center lies on [tex]AB[/tex].
Only one circle passes through [tex]C[/tex] and [tex]L[/tex] with center on [tex]AB[/tex] (its center is the intersection of [tex]AB[/tex] with the perpendicular bisector of [tex]CL[/tex]), so it is this circle: [tex]r=\frac{LL'}{2}=\frac{15}{2}[/tex].