Circles: Chords and Tangents: Problems with Solutions

Solution:
Let [tex]M[/tex] be the foot of the perpendicular from [tex]O[/tex] to the chord: [tex]OM \perp AB[/tex]. The perpendicular from the center bisects the chord, so [tex]AM=\frac{AB}{2}=8[/tex].
From the right [tex]\triangle OAM[/tex]: [tex]OM=\sqrt{OA^2-AM^2}=\sqrt{100-64}=6[/tex].

Solution:
Let [tex]AM=3k;\ BM=2k[/tex]. By the intersecting chords theorem [tex]AM \times BM=CM \times DM[/tex], so [tex]6k^2=96[/tex] and [tex]k=4[/tex].
Hence [tex]AB=5k=20[/tex].

Solution:
[tex]BA=BD+DA=26[/tex]. By the tangent–secant theorem [tex]BC^2=BD \times BA=8 \times 26=208[/tex], so [tex]BC=\sqrt{208}=4\sqrt{13}[/tex].

Solution:
The distance between the centers is [tex]O_1O_2=50+32=82[/tex]. The radii to the points of tangency are perpendicular to the tangent; a line through the smaller center parallel to the tangent gives a right triangle with hypotenuse [tex]O_1O_2[/tex] and a leg equal to the difference of the radii, so [tex]T_1T_2=\sqrt{O_1O_2^2-(50-32)^2}=\sqrt{82^2-18^2}=\sqrt{6400}=80[/tex].

Solution:
The line of centers is the perpendicular bisector of the common chord, and half the chord is [tex]\frac{24}{2}=12[/tex]. The distances from the centers to the chord are [tex]d_1=\sqrt{13^2-12^2}=5[/tex] and [tex]d_2=\sqrt{15^2-12^2}=9[/tex].
The centers lie on opposite sides of the chord, so [tex]O_1O_2=5+9=14[/tex].

Solution:
Let [tex]M[/tex] and [tex]N[/tex] be the feet of the perpendiculars from [tex]O[/tex]: [tex]OM \perp AB;\ ON \perp CD[/tex], and [tex]OM=ON[/tex]. The perpendicular from the center bisects a chord, so [tex]AM=\frac{AB}{2};\ CN=\frac{CD}{2}[/tex].
The right triangles have [tex]OA=OC[/tex] (radii) and [tex]OM=ON[/tex], so [tex]\triangle OAM \cong \triangle OCN[/tex] and [tex]AM=CN[/tex].
Hence [tex]AB=CD[/tex].

Solution:
The angle subtending the diameter [tex]AB[/tex] is right: [tex]\angle AMB=90^\circ[/tex], so [tex]BM \perp AM[/tex].
Since [tex]AM \parallel BN[/tex], also [tex]BM \perp BN[/tex], i.e. [tex]\angle MBN=90^\circ[/tex].
So the inscribed angle [tex]\angle MBN=90^\circ[/tex] subtends the chord [tex]MN[/tex], hence [tex]MN[/tex] is a diameter.

Solution:
[tex]O_1A=O_1B[/tex] (radii of [tex]k_1[/tex]), so [tex]O_1[/tex] lies on the perpendicular bisector of [tex]AB[/tex].
Likewise [tex]O_2A=O_2B[/tex], so [tex]O_2[/tex] also lies on it.
Two different points determine a line, so [tex]O_1O_2[/tex] is the perpendicular bisector of [tex]AB[/tex].

Solution:
Let [tex]R=5k;\ r=3k[/tex]. For internally tangent circles the distance between the centers equals the difference of the radii: [tex]O_1O_2=R-r=2k[/tex].
So [tex]2k=14[/tex], [tex]k=7[/tex], and [tex]R=35;\ r=21[/tex].

Solution:
The diameter is [tex]8+3=11[/tex], so [tex]R=\frac{11}{2}[/tex].
The chord meets the diameter at [tex]M[/tex], which is at distance [tex]3[/tex] from one end of the diameter, so [tex]OM=R-3=\frac{5}{2}[/tex].

Solution:
[tex]r_1+r_2=5+3=8[/tex], so [tex]O_1O_2=r_1+r_2[/tex]: the circles are tangent externally.
Externally tangent circles have two common external tangents and one common tangent at the point of contact, [tex]3[/tex] in total.

Solution:
Drop the perpendiculars [tex]AA_1 \perp a;\ BB_1 \perp a;\ OO_1 \perp a[/tex]. Then [tex]AA_1B_1B[/tex] is a right trapezoid and [tex]OO_1[/tex] is its midsegment, since [tex]O[/tex] is the midpoint of [tex]AB[/tex].
So [tex]OO_1=\frac{AA_1+BB_1}{2}=\frac{3+11}{2}=7[/tex].

Solution:
Since [tex]AM \perp MB[/tex], [tex]AB[/tex] is a diameter and [tex]O[/tex] is its midpoint. Let [tex]K[/tex] and [tex]L[/tex] be the midpoints of [tex]AM[/tex] and [tex]MB[/tex]; [tex]OK;\ OL[/tex] are midsegments of [tex]\triangle AMB[/tex], so [tex]OK \parallel MB;\ OL \parallel AM[/tex], and they are perpendicular to the chords.
Hence [tex]OK=\frac{MB}{2}=2;\ OL=\frac{AM}{2}=\frac{3}{2}[/tex].

Solution:
The angle between a tangent and a chord equals the inscribed angle on the same arc. For the chord [tex]BC[/tex]: [tex]\angle PBC=\angle BAC=60^\circ[/tex] and [tex]\angle PCB=\angle BAC=60^\circ[/tex].
Then [tex]\angle BPC=60^\circ[/tex], so all angles of [tex]\triangle BPC[/tex] are equal and it is equilateral.

Solution:
The reflection in the line of centers [tex]O_1O_2[/tex] swaps the two common external tangents, so it swaps the points of tangency on each circle. Hence [tex]O_1O_2[/tex] is the perpendicular bisector of the segments joining them: [tex]T_1T_1' \perp O_1O_2[/tex] and [tex]T_2T_2' \perp O_1O_2[/tex].
So [tex]T_1T_1' \parallel T_2T_2'[/tex], and since the radii are different these segments have different lengths; therefore [tex]T_1T_1'T_2'T_2[/tex] is a trapezoid (isosceles, since [tex]T_1T_2=T_1'T_2'[/tex]).

Solution:
Let [tex]H[/tex] be the foot of the perpendicular from [tex]O[/tex] to [tex]l[/tex]: [tex]OH \perp l[/tex]. The perpendicular from the center bisects every chord on [tex]l[/tex], so [tex]HA=HB[/tex] and [tex]HC=HD[/tex].
Hence [tex]AC=HA-HC=HB-HD=BD[/tex].

Solution:
Compare the radius with the distance [tex]d=8[/tex] from the center to the line. For [tex]R=6 \lt d[/tex] the line does not meet the circle: [tex]0[/tex] common points; for [tex]R=8=d[/tex] it is tangent: [tex]1[/tex] common point; for [tex]R=10 \gt d[/tex] it is a secant: [tex]2[/tex] common points.

Solution:
The radius to the point of tangency is perpendicular to the tangent: [tex]OA \perp CA[/tex], so [tex]CA=\sqrt{OC^2-OA^2}=\sqrt{16-4}=2\sqrt{3}[/tex]. Tangent segments from a point are equal: [tex]CA=CB[/tex].
Also [tex]\sin\angle OCA=\frac{OA}{OC}=\frac{1}{2}[/tex], so [tex]\angle OCA=30^\circ;\ \angle ACB=60^\circ[/tex]; an isosceles triangle with this vertex angle is equilateral: [tex]AB=CA=CB=2\sqrt{3}[/tex].

Solution:
In one hour the tip goes around a full circle with radius [tex]10[/tex] cm: [tex]C=2\pi r=2\pi \times 10=20\pi[/tex] cm.
In [tex]10[/tex] hours: [tex]10 \times 20\pi=200\pi[/tex] cm.