Circles: Chords and Tangents: Difficult Problems with Solutions

Solution:
The shortest distance is measured along the line through the center [tex]O[/tex], so [tex]OA=R+25[/tex]. The radius to the point of tangency [tex]T[/tex] is perpendicular to the tangent: [tex]\angle OTA=90^\circ[/tex], [tex]OT=R[/tex].
By the Pythagorean theorem [tex](R+25)^2=R^2+35^2[/tex], i.e. [tex]50R+625=1225[/tex], so [tex]R=12[/tex].

Solution:
[tex]R=\frac{6+14}{2}=10[/tex], and the distance from the center [tex]O[/tex] to [tex]P[/tex] is [tex]OP=10-6=4[/tex].
The distance from [tex]O[/tex] to the chord is [tex]d=OP\sin 45^\circ=2\sqrt{2}[/tex].
[tex]\frac{AB}{2}=\sqrt{R^2-d^2}=\sqrt{100-8}=\sqrt{92}=2\sqrt{23}[/tex], so [tex]AB=4\sqrt{23}[/tex].

Solution:
The radius to the point of tangency is perpendicular to the tangent: [tex]OB \perp t[/tex], so [tex]OB \perp AC[/tex]. A diameter perpendicular to a chord bisects the chord and its arc, so [tex]BA=BC=10[/tex].
Let [tex]H[/tex] be the midpoint of [tex]AC[/tex]; then [tex]BH=\sqrt{10^2-6^2}=8[/tex].
[tex]R=\frac{AB \times BC}{2 \times BH}=\frac{100}{16}=\frac{25}{4}[/tex].

Solution:
The centers are [tex]O_1O_2=3+1=4[/tex] apart. The common internal tangent passes through the point of contact and is perpendicular to [tex]O_1O_2[/tex].
The external tangent forms an angle [tex]\varphi[/tex] with [tex]O_1O_2[/tex], where [tex]\sin\varphi=\frac{3-1}{4}=\frac{1}{2}[/tex] (difference of the radii over the distance between the centers), so [tex]\varphi=30^\circ[/tex].
Hence the angle between the two tangents is [tex]90^\circ-30^\circ=60^\circ[/tex].

Solution:
The angle between the tangent and the chord [tex]TA[/tex] equals the inscribed angle on the same arc: [tex]\angle MTA=\angle MBT[/tex].
The triangles also share [tex]\angle M[/tex], so [tex]\triangle MTA \sim \triangle MBT[/tex].
Hence [tex]\frac{MT}{MB}=\frac{MA}{MT}[/tex], i.e. [tex]MT^2=MA \times MB[/tex].

Solution:
[tex]BC=8[/tex]. The angle between the tangent and the chord [tex]CA[/tex] equals the inscribed angle on the same arc: [tex]\angle MCA=\angle MBC[/tex]; with the common angle at [tex]M[/tex], [tex]\triangle MCA \sim \triangle MBC[/tex].
So [tex]\frac{MC}{MB}=\frac{MA}{MC}=\frac{AC}{BC}=\frac{3}{4}[/tex]. [tex]M[/tex] lies beyond [tex]A[/tex]; let [tex]MA=x[/tex], then [tex]MB=x+10[/tex].
From the ratios [tex]x=\frac{3}{4}MC=\frac{9}{16}(x+10)[/tex], so [tex]x=\frac{90}{7}[/tex] and [tex]MC=\frac{4}{3}x=\frac{120}{7}[/tex].

Solution:
Let [tex]\theta[/tex] be the angle between the chord and the tangent at [tex]A[/tex]; it equals the inscribed angle on the arc [tex]AB[/tex], so [tex]AB=2R\sin\theta[/tex]. The distance from [tex]B[/tex] to the tangent is [tex]d=AB\sin\theta=\frac{AB^2}{2R}[/tex].
So [tex]\frac{AB^2}{4}+AB=3[/tex], i.e. [tex]AB^2+4AB-12=0[/tex], and the positive root is [tex]AB=2[/tex].

Solution:
The product [tex]AB \times AC[/tex] is the same for every secant through [tex]A[/tex]; for the secant through the center it equals [tex](9-7)(9+7)=32[/tex].
Since [tex]AB=BC[/tex], [tex]AC=2AB[/tex], so [tex]2AB^2=32[/tex] and [tex]AB=4[/tex].

Solution:
Let [tex]MC=mk;\ MD=nk[/tex]. The diameter perpendicular to the chord bisects it: [tex]AM=MB=\frac{a}{2}[/tex].
By the intersecting chords theorem [tex]AM \times MB=MC \times MD[/tex], i.e. [tex]\frac{a^2}{4}=mnk^2[/tex], so [tex]k=\frac{a}{2\sqrt{mn}}[/tex].
[tex]2R=(m+n)k[/tex], hence [tex]R=\frac{(m+n)a}{4\sqrt{mn}}[/tex].

Solution:
The central angle on the arc is [tex]\angle AOB=2\alpha[/tex]; in the isosceles [tex]\triangle AOB[/tex] with legs [tex]R[/tex]: [tex]AB=2R\sin\alpha[/tex].
[tex]M[/tex] halves the arc, so [tex]\angle AOM=\alpha[/tex], and in the same way [tex]AM=2R\sin\frac{\alpha}{2}[/tex].

Solution:
Let [tex]P[/tex] be the point of the arch above [tex]C[/tex]; the height is [tex]CP[/tex]. [tex]\triangle APB[/tex] is right because it subtends the diameter, and [tex]CP[/tex] is its altitude to the hypotenuse.
Since [tex]CB=5[/tex], [tex]CP[/tex]² = [tex]AC \times CB=1 \times 5=5[/tex], so [tex]h=\sqrt{5}[/tex] ft.

Solution:
[tex]\angle AMD=\angle CMB[/tex] (vertical angles) and [tex]\angle DAB=\angle DCB[/tex] (inscribed angles on the arc [tex]DB[/tex]), so [tex]\triangle AMD \sim \triangle CMB[/tex] with ratio [tex]\frac{AM}{CM}=\frac{4}{3}[/tex].
The areas of similar triangles are in the ratio of the squares of corresponding sides, so [tex]S_{CMB}=S_{AMD} \times \left(\frac{3}{4}\right)^2=\frac{9}{8}[/tex].

Solution:
The angle between the tangent [tex]BC[/tex] and the chord [tex]CD[/tex] equals the inscribed angle on the same arc: [tex]\angle BCD=\angle BAC[/tex]; with the common angle at [tex]B[/tex], [tex]\triangle BCD \sim \triangle BAC[/tex].
So [tex]\frac{BC}{BA}=\frac{CD}{AC}=\frac{BD}{BC}[/tex]: [tex]\frac{12}{BA}=\frac{6}{9}[/tex], hence [tex]BA=18[/tex], and [tex]BD=\frac{BC^2}{BA}=\frac{144}{18}=8[/tex].

Solution:
By the theorem on two secants [tex]MA \times MB=MD \times MC[/tex]: [tex]3MB=36[/tex], so [tex]MB=12[/tex] and [tex]AB=MB-MA=9[/tex].
The triangles have a common angle at [tex]M[/tex] and proportional sides [tex]\frac{MA}{MC}=\frac{MD}{MB}=\frac{1}{3}[/tex], so [tex]\triangle MAD \sim \triangle MCB[/tex]; hence [tex]\frac{AD}{CB}=\frac{1}{3}[/tex] and [tex]BC=6[/tex].

Solution:
Since [tex]\angle AA_1B=\angle AB_1B=90^\circ[/tex], the points [tex]A;\ B;\ A_1;\ B_1[/tex] lie on the circle with diameter [tex]AB[/tex].
[tex]AA_1;\ BB_1[/tex] are two chords of this circle meeting at [tex]H[/tex], so by the intersecting chords theorem [tex]AH \times HA_1=BH \times HB_1[/tex].
In the same way (the circle with diameter [tex]BC[/tex]) [tex]BH \times HB_1=CH \times HC_1[/tex].

Solution:
Let [tex]O[/tex] be the intersection point of the diagonals; it bisects them. [tex]AC=81[/tex], so [tex]AO=\frac{81}{2}[/tex] and [tex]OL=65-\frac{81}{2}=\frac{49}{2}[/tex].
By the intersecting chords theorem for the chords [tex]AL[/tex] and [tex]BD[/tex]: [tex]OB \times OD=OA \times OL[/tex], i.e. [tex]OB^2=\frac{81}{2} \times \frac{49}{2}[/tex], so [tex]OB=\frac{63}{2}[/tex].
Then [tex]BD=63[/tex] and [tex]AB=BD=63[/tex].

Solution:
Since [tex]\angle BAD=\angle DAC=60^\circ[/tex], the chords subtending equal inscribed angles are equal: [tex]BD=CD[/tex]. The quadrilateral [tex]ABDC[/tex] is cyclic, so [tex]\angle BDC=180^\circ-120^\circ=60^\circ[/tex], hence [tex]\triangle BCD[/tex] is equilateral: [tex]BC=BD=CD[/tex].
By Ptolemy's theorem [tex]AD \times BC=AB \times CD+AC \times BD[/tex]; dividing by [tex]BC[/tex] gives [tex]AD=AB+AC=3[/tex].

Solution:
For the external tangent, a line through the smaller center parallel to the tangent gives a right triangle with hypotenuse equal to the distance between the centers and a leg equal to the difference of the radii: [tex]\sqrt{25^2-(11-4)^2}=\sqrt{625-49}=24[/tex].
For the internal tangent the leg is the sum of the radii: [tex]\sqrt{25^2-(11+4)^2}=\sqrt{625-225}=20[/tex].

Solution:
Let [tex]\overset{\frown}{BC}=3k;\ \overset{\frown}{AC}=5k[/tex]. An angle with vertex outside the circle equals half the difference of the arcs between its sides: [tex]\angle AMC=\frac{\overset{\frown}{AC}-\overset{\frown}{BC}}{2}[/tex].
So [tex]\frac{5k-3k}{2}=32^\circ[/tex], [tex]k=32^\circ[/tex], and [tex]\overset{\frown}{BC}=96^\circ;\ \overset{\frown}{AC}=160^\circ[/tex].

Solution:
Let [tex]O[/tex] be the center and [tex]CH[/tex] the altitude; [tex]AH=3[/tex] and [tex]CH=4[/tex]. The radius to the point of tangency is perpendicular to the tangent: [tex]OA \perp AC[/tex], so [tex]\angle OAH=\angle ACH[/tex].
The right triangles give [tex]\triangle OAH \sim \triangle ACH[/tex], so [tex]\frac{OA}{AC}=\frac{AH}{CH}[/tex] and [tex]OA=\frac{5 \times 3}{4}=\frac{15}{4}[/tex].

Solution:
[tex]O_1T_1T_2O_2[/tex] is a right trapezoid whose parallel sides are the radii [tex]3[/tex] and [tex]9[/tex] (they are perpendicular to the tangent). The distance [tex]AK[/tex] from [tex]A[/tex] to the tangent is the segment through [tex]A[/tex] parallel to them, and [tex]A[/tex] divides [tex]O_1O_2[/tex] in the ratio [tex]1:3[/tex], since [tex]\frac{O_1A}{O_1O_2}=\frac{3}{12}=\frac{1}{4}[/tex].
So [tex]AK=3+\frac{1}{4}(9-3)=\frac{9}{2}[/tex].

Solution:
For [tex]k_1[/tex], the tangent from [tex]N[/tex] and the secant through [tex]A[/tex] and [tex]B[/tex] give [tex]NT_1^2=NA \times NB[/tex]. For [tex]k_2[/tex] in the same way [tex]NT_2^2=NA \times NB[/tex].
Hence the squares are equal and [tex]NT_1=NT_2[/tex]: the line [tex]AB[/tex] bisects the segment of the common tangent.

Solution:
The diameter through [tex]C[/tex] is divided by [tex]C[/tex] into segments with product [tex](7-3)(7+3)=40[/tex]; by the intersecting chords theorem also [tex]AC \times CB=40[/tex].
Let [tex]AC=x[/tex]; since [tex]AC+CB=13[/tex], [tex]x(13-x)=40[/tex], i.e. [tex]x^2-13x+40=0[/tex], so [tex]x_1=5;\ x_2=8[/tex].
The chord is divided into segments [tex]5;\ 8[/tex].

Solution:
The radii to the points of tangency are perpendicular to the tangent, so [tex]O_1T_1 \parallel O_2T_2[/tex] and [tex]\triangle PO_1T_1 \sim \triangle PO_2T_2[/tex], hence [tex]\frac{PO_1}{PO_2}=\frac{R}{r}[/tex].
[tex]P[/tex] lies on the side of the smaller circle, so [tex]PO_1=PO_2+R+r[/tex]. Then [tex]\frac{PO_2+R+r}{PO_2}=\frac{R}{r}[/tex], which gives [tex]PO_2=\frac{r(R+r)}{R-r}[/tex].

Solution:
Let the common tangent at [tex]M[/tex] meet [tex]AB[/tex] at [tex]N[/tex]. Tangent segments from a point are equal: [tex]NA=NM=NB[/tex].
So [tex]M[/tex] lies on the circle with diameter [tex]AB[/tex], hence [tex]\angle AMB=90^\circ[/tex].
The length of the common external tangent of two touching circles is twice the geometric mean of the radii: [tex]AB=2\sqrt{4 \times 1}=4[/tex].

Solution:
The angle between the tangent [tex]AD[/tex] and the chord [tex]AC[/tex] equals the inscribed angle on the same arc: [tex]\angle DAC=\angle ABC[/tex]. Also [tex]\angle DCA=\angle CAB[/tex] (alternate angles).
So [tex]\triangle DAC \sim \triangle CBA[/tex], hence [tex]\frac{DC}{CA}=\frac{CA}{AB}[/tex], i.e. [tex]AC^2=ab[/tex] and [tex]AC=\sqrt{ab}[/tex].

Solution:
The side of the square is [tex]25[/tex]. Place [tex]A(0,0),\ B(25,0),\ C(25,25),\ D(0,25)[/tex]; the center is [tex]Q(r,r)[/tex] and the radius is [tex]r[/tex]. The circle meets [tex]BC[/tex] at the point [tex](25,2)[/tex], so [tex](25-r)^2+(2-r)^2=r^2[/tex], i.e. [tex]r^2-54r+629=0[/tex], and [tex]r_1=17;\ r_2=37[/tex].
The radius is less than the side, [tex]r<25[/tex], so [tex]r=17[/tex].

Solution:
The line [tex]AO[/tex] passes through the center and is perpendicular to [tex]BC=6[/tex], so it bisects it at [tex]H[/tex]; hence [tex]AC=AB=5[/tex].
[tex]AH=\sqrt{25-9}=4[/tex]. The diameter [tex]AD[/tex] through [tex]O[/tex] gives the right [tex]\triangle ABD[/tex] with altitude [tex]AH[/tex], so [tex]AB^2=AH \times AD[/tex], and [tex]AD=\frac{25}{4};\ R=\frac{25}{8}[/tex].

Solution:
The points [tex]O;\ O_1;\ T[/tex] are collinear (the line of centers passes through the point of contact), so [tex]\angle OTA=\angle O_1TA_1[/tex].
[tex]\triangle TOA[/tex] and [tex]\triangle TO_1A_1[/tex] are isosceles: [tex]\angle OAT=\angle OTA[/tex], [tex]\angle O_1A_1T=\angle O_1TA_1[/tex]. Hence [tex]\angle OAT=\angle O_1A_1T[/tex]; these are corresponding angles, so [tex]OA \parallel O_1A_1[/tex].
Also [tex]\triangle TOA \sim \triangle TO_1A_1[/tex] with ratio equal to the ratio of the radii, so [tex]\frac{TA}{TA_1}=\frac{R}{r}[/tex].

Solution:
The centers of the circles are the midpoints [tex]M[/tex] and [tex]N[/tex] of the legs, so the distance between the centers is the midsegment: [tex]MN=\frac{AB+CD}{2}[/tex].
The sum of the radii is [tex]\frac{AD}{2}+\frac{BC}{2}=\frac{AD+BC}{2}[/tex], and by the condition it equals [tex]MN=\frac{AB+CD}{2}[/tex].
So [tex]MN=\frac{AD}{2}+\frac{BC}{2}[/tex]: the distance between the centers equals the sum of the radii, hence the circles are tangent externally.

Solution:
Let [tex]AB=CD=a[/tex]. By the theorem on two secants [tex]PA \times PB=PC \times PD[/tex], where [tex]PA=PB+a[/tex] and [tex]PC=PD+a[/tex].
So [tex](PB+a)PB=(PD+a)PD[/tex], i.e. [tex](PB-PD)(PB+PD+a)=0[/tex]. The second factor is positive, hence [tex]PB=PD[/tex], and then [tex]PA=PC[/tex].

Solution:
The center lies on the angle bisector: [tex]\angle ABO=15^\circ[/tex]. In the right [tex]\triangle ABO[/tex] (the radius [tex]OA[/tex] is perpendicular to the tangent): [tex]AB=BO\cos 15^\circ[/tex]. The tangent segments are equal, [tex]BA=BC[/tex], so [tex]AC \perp BO[/tex]; let [tex]H[/tex] be the intersection point of [tex]AC[/tex] and [tex]BO[/tex].
[tex]AH=AB\sin 15^\circ=14\sin 15^\circ\cos 15^\circ=7\sin 30^\circ=\frac{7}{2}[/tex], so [tex]AC=2AH=7[/tex].

Solution:
The angle between the tangent and the chord [tex]CB[/tex] equals the inscribed angle on it: [tex]\angle PCB=\angle CAB[/tex]. Then [tex]\angle PCL=\angle PCB+\angle BCL[/tex], and [tex]\angle PLC=\angle CAL+\angle ACL[/tex] as an exterior angle of [tex]\triangle ALC[/tex].
Since [tex]\angle BCL=\angle ACL[/tex] and [tex]\angle PCB=\angle CAB[/tex], [tex]\angle PCL=\angle PLC[/tex], so [tex]\triangle PCL[/tex] is isosceles and [tex]PC=PL[/tex].

Solution:
Let [tex]I[/tex] be the intersection point of the segment [tex]AO[/tex] with the circle. By symmetry, [tex]AO[/tex] bisects [tex]\angle BAC[/tex] and [tex]I[/tex] is the midpoint of the arc cut off by [tex]BC[/tex]: [tex]\overset{\frown}{BI}=\overset{\frown}{IC}[/tex].
The angle between the tangent [tex]AB[/tex] and the chord [tex]BI[/tex] equals the inscribed angle on it: [tex]\angle ABI=\angle BCI[/tex], and equal arcs give [tex]\angle BCI=\angle IBC[/tex].
Hence [tex]\angle ABI=\angle IBC[/tex], i.e. [tex]BI[/tex] bisects the angle at [tex]B[/tex]. So [tex]I[/tex] lies on two angle bisectors of [tex]\triangle ABC[/tex] and is its incenter.