Problems involving circles are among the most interesting in the mathematics. To learn these problems one should solve it independentlu, not looking even at the given solutins.

**Circle** is the set of points in plane on the same distance from the fixed point. The fixed point is named **centre**, and the fixed distance is **radius**.

**Chord** is the line joining two points on a circle. When the line passes through the centre, the chord is **diameter**.

**Tangent** is a straight line which touches a curve at a point. A tangent to a circle is at right angles to the radius of the circle at its contact.

When a straight line is drawn joining two points on the **circumference** of a circle it divides it into two parts, the **minor segment** and the **major segment**.

The region of a circle formed by two **radii** and part of the circle's circumference is called a **sector** of the circle.

The smaller region formed is called the **minor** sector, the larger region formed is called the **major** sector.

**Perimeter** of the shape is the name given to the boundary of the shape. It is also sometimes used to mean the length of the boundary.

**Inscribe** means to draw inside. When one shape fits inside another it is said to be inscribed. For example, a **circle** can be inscribed in a **triangle** to touch the sides, and a **square** can be inscribed in a circle so that its **vertices** (a plural of vertex) are on the circle. The **vertex** is the intersection of two sides of a plane figure.

**Semi** means 'half'. A **semi-circle** is half of a circle formed by cutting along a diameter.

A tangent to a circle is at **right angle** to the radius of the circle at its contact.

If *A*, *B* and *C* are points on a circle where the line *AC* is a diameter of the circle, then the angle *ABC* is a right angle.

**1.** Let *k* be an outside inscribed circle that connects the side *BC* and extensions of *BA* and *CA* of the triangle *ABC*. Prove that tangent from vertex *A* to the connect point of *k* is equal to semiperimeter of the triangle.

**Solution**

On the graph above are tangents long from *A* to circle *k* in points *M* and *N*. Tangent lines *AM* and *AN* are equal. Also, *BM* = *BN* and *CP* = *CN*.

Hence, perimeter of the circle is *AB* + *BC* + *CA* = *AB* + (*BM* + *CP*) + *AC* = (*AB* + *BM*) + (*CP* + *AC*) = 2×*AM*, ie. semi perimeter = *AM*.♦

**2.** Let *s* be a semi perimeter and *c* hypotenuses of given rectangular triangle, than the radius of the inscribed circle is *s* – *c*.

**Solution**

Projections of the center of (inscribed) circle to sides *a*, *b* and *c* are successively *M*, *N* and *P* (draw!). That are points on sides *BC*, *CA* and *AB* of the given rectangular triangle, with right angle at the point *C*, and semi radius of the circle *r* = *CM* = *CN*. Tangent lines, drown from points of the triangle to the inscribed circle are equal. So the perimeter of the triangle is 2*s* = *AB* + *BC* + *CA* = 2*AP* + 2*BP* +2*r* = 2*c* + 2*r*. Hence *s* = *c* + *r*, i.e. *r* = *c* – *r*.♦

**3.** Circles *k*_{1} and *k*_{2} have contact in point *A*. Chord *a*, containing point *A*, cuts these circles respectively at points *P*_{1} and *P*_{2}. Prove that tangents of given circles in points *P*_{1} and *P*_{2} are parallel.

**Solution**

Triangle *P*_{1}*AO*_{1} is isosceles, as triangle *P*_{2}*AO*_{2} too. Consequently, all four angles *AP*_{1}, *P*_{1}*AO*_{1} ..., are equal, so the angles between tangents and given chords are equal. Meaning, the tangents are parallel.♦

**4.** Circles *k*_{1} and *k*_{2} have contact outside in point *A*. Their common tangent connect the circle *k*_{1} in point *B*, the circle *k*_{2} in point *C*. Prove, the angle *BAC* is rectangular.

**Solution**

Let *t*_{1} and *t*_{2} be common tangents of given circles (draw!). Intersection of the tangents is point *S*. From *SA* = *SB* = *SC* follows that *S* is center of the circle passing by points *A*, *B*, *C*, whereas the *BC* is diameter. Meaning, the angle *BAC* is rectangular.♦

**5.** Given are the circle *k*(*O*, *r*) and its tangent *a* with connection point *A*. The tangent *a* has points *B* and *C* with tangents line *BD* and *CE* ending on the circle *k*. Prove that angles *BOC* and *DAE* are equal or **supplementary** (whose sum is 180°).

**Solution**

Chord *AD* is perpendicular to line *OB* (of radius) in point *F*, graph above. Rectangular triangles *BOA* and *BAF* have common angle in point *B*, so they have all tree corresponding angles the same. Thus, angle *BOA* is **complementary** (the sum of the measures of two angles is 90°) with angle *DAO*, so *BOC* and *DAE* are supplementary angles.

**6.** Let *AP* and *AQ* be tangent lines on the circle *k*(*O*, *r*). If through any point *M* of the circle in major segment *PQ* a tangent that passes lines *AP* and *AQ* successively in points *B* and *C*, prove that *AB* + *AC* – *BC* and angle *BOC* are constant.

**Solution**

First, *AB* + *AC* – *BC* = (*x* + *y*) + (*x* + *z*) - (*y* + *z*) = 2*x* = const. Second, ∠*CBA* + ∠*ACB* = 180° - α and ∠*MOB* + ∠*CBA*/2 = 90°. Thus ∠*BOM* = 90° - ∠*CBA*/2, ∠*MOC* = 90° - ∠*ACB*/2, so ∠*BOC* = ∠*BOM* + ∠*MOC* = (90° - ∠*CBA*/2) + (90° - ∠*ACB*/2) = 180° - (∠*CBA* + ∠*ACB*)/2 = 180° - (90° - α/2) = 90° + α/2 = const. for given α = ∠*BAC*.♦

**7.** Given are circle *k*, its diameter *AB* and point *P* near circle beside line *AB*. Using only ruler, construct perpendicular line through *P* to line *AB*.

**Solution**

If the point *P* is outside the circle, lines *AP* and *BP* intersects the circle in points *M* and *N*. Then *AN*⊥*BP* and *BM*⊥*AP*. Let the intersection of lines *AN* and *BM* be point *Q*. Line *PQ* is required normal (line *n* on drawing above), the third height of a triangle *ABP*. If the point *P* is inside circle, points *P* and *Q* change its roles.♦

**8.** Given angle 19° divide to 19 equal parts using only pair of compasses.

**Solution**

Draw by compasses a circle with center in top of the given angle. Increase, transfer given angle 19 times. Forasmuch as 19×19 = 361 we have central angle 1°. Adding obtained angle into given angle, we have requested parts.♦

Translated from Serbian: Krugovi

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