Lecture Notes on Geometry

Subsection 3.1 Lines and circles

1. From the equation of the line get \(x\) or \(y\) (try to select the variable that makes the computation easier),
2. substitute this variable into the equation of the circle and solve the quadratic equation,
3. substitute separately the value(s) obtained at the previous step into the linear equation in step 1 to obtain the value(s) of the other variable.

Subsection 3.2 Exercises

Subsection 3.3 Determining equation of a circle

Subsection 3.4 Tangent to a circle

Subsection 3.5 Exercises

Subsection 3.6 Touching circles

• If \(r_1+r_2=d,\) then the circles touch externally.

  

• If (suppose \(r_1 > r_2\)) \(r_1-r_2=d,\) then the circles touch internally.

  

Subsection 3.7 Exercises

Subsection 3.8 Common chord, common tangent

• The system of equations \(C_1-C_2=0, C_1=0\) has no solution. It means that we have two disjoint circles.

  

• The system of equations \(C_1-C_2=0, C_1=0\) has one solution. In this case we have touching circles, the line is a common tangent line.

  

• The system of equations \(C_1-C_2=0, C_1=0\) has two solutions. The line is a common chord, the two circles intersect in two points.

  

Subsection 3.9 Exercises

Subsection 3.10 Common tangents to two circles

• There is a very special degenerate case. If \(d=0\) and \(r_1=r_2,\) then the two circles coincide and therefore there are infinitely many common tangent lines.
• If \(r_2 > r_1+d,\) then there is no common tangent line.
• If \(r_2=r_1+d,\) then the two circle touch internally and there is one common tangent line.
• If \(r_2 < r_1+d < r_2+2r_1,\) then there are two tangent lines in common.
• If \(d=r_1+r_2,\) then the two circles touch externally and there are three common tangent lines.
• If \(d > r_1+r_2,\) then there are four common tangent lines.

• If two circles of radius \(r_{1}\) and \(r_{2}\) touch each other externally, then the length of the direct common tangent is \(\sqrt{4r_{1}r_{2}}.\)

  

  We would like to determine the length of \(PQ.\) Draw a line \(OR\) parallel to \(PQ.\) Here we have that \(\angle OPQ=90^{\circ}, \angle O'QP=90^{\circ}.\) Since opposite sides are parallel and interior angles are \(90^{\circ}\text{,}\) therefore \(OPQR\) is a rectangle. So \(OP = QR = r_{1}\) and \(PQ = OR.\) Hence \(\angle ORO'=90^{\circ}.\) We apply the theorem of Pythagoras:
  \begin{equation*} (r_1+r_2)^2=OO'^2=OR^2+O'R^2=PQ^2+(r_1-r_2)^2\longrightarrow 4r_1r_2=PQ^2 \longrightarrow PQ=\sqrt{4r_1r_2}. \end{equation*}
• There are two circle of radius \(r_{1}\) and \(r_{2}\) which intersect each other at two points. If their centers are \(d\) units apart , then the length of the direct common tangent between them is \(\sqrt{d^{2}-(r_{1}-r_{2})^{2}}.\)

  

  The computation follows the same lines as in the previous case, however here we have \(OO'=d.\) Hence we obtain
  \begin{equation*} d^2=PQ^2+(r_1-r_2)^2\longrightarrow PQ^2=d^2-(r_1-r_2)^2 \longrightarrow PQ=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}. \end{equation*}
• If the centers of two circle of radius \(r_{1}\) and \(r_{2}\) are \(d\) units apart , then the length of the direct common tangent between them is \(\sqrt{d^{2}-(r_{1}-r_{2})^{2}}.\)

  

  The computation is similar to the one applied in the previous case.
• If the centers of two circle of radius \(r_{1}\) and \(r_{2}\) are \(d\) units apart , then the length of the transverse common tangent between them is \(\sqrt{d^{2}-(r_{1}+r_{2})^{2}}.\)

  

  Since \(PQ\) is parallel to \(RO'\) the triangle \(ORO'\) is a right triangle and we may apply the theorem of Pythagoras:
  \begin{equation*} d^2=OO'^2=OR^2+O'R^2=(r_1+r_2)^2+PQ^2 \longrightarrow PQ=\sqrt{d^2-(r_1+r_2)^2}. \end{equation*}

Subsection 3.11 Exercises

Subsection 3.12 Angle properties of the circle

Theorem 3.66. (A).

The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.

Join points \(P\) and \(O\) and extend the line through \(O\) as shown in the diagram.

Note that \(AO = BO = PO = r\) the radius of the circle. Therefore triangles \(PAO\) and \(PBO\) are isosceles. Let \(\angle APO = \angle PAO = a^{\circ}\) and \(\angle BPO = \angle PBO = b^{\circ}\) Then angle \(AOX\) is \(2a^{\circ}\) (exterior angle of a triangle) and angle \(BOX\) is \(2b^{\circ}\) (exterior angle of a triangle). Therefore we have

\begin{equation*} \angle AOB = 2a^{\circ} + 2b^{\circ} = 2(a + b)^{\circ} = 2\angle APB. \end{equation*}

Let \(\angle AXB = x^{\circ}\) and \(\angle AYB = y^{\circ}\) Then by Theorem A we obtain \(\angle AOB = 2x^{\circ} = 2y^{\circ}\) Therefore \(x = y.\)

Application of Theorem A provides the following figure:

We have that \(2x+2y=360^{\circ}.\) Thus we get that \(x+y=180^{\circ}.\)

Let \(\angle STQ = x^{\circ}, \angle RTS = y^{\circ}\) and \(\angle TRS = z^{\circ}\) where \(RT\) is a diameter. Then \(\angle RST = 90^{\circ}\) (Theorem of Thales, angle subtended by a diameter). Also \(\angle RTQ = 90^{\circ}\) (tangent is perpendicular to radius). Hence \(x + y = 90\) and \(y + z = 90.\) Therefore \(x = z.\) But \(\angle TXS\) is in the same segment as \(\angle TRS\) and therefore \(\angle TXS = x^{\circ}.\)

Subsection 3.13 Exercises

Subsection 3.14 Chords in circles

If the intersection point is inside the circle.

Consider triangles \(APC\) and \(BPD.\) We have that \(\angle APC = \angle BPD\) (vertically opposite) and \(\angle CDB = \angle CAB\) (angles in the same segment), \(\angle ACD = \angle DBA\) (angles in the same segment). Therefore triangle \(CAP\) is similar to triangle \(BDP.\) Therefore \(AP/PD=CP/PB\) and \(AP\cdot PB = CP\cdot PD,\) which can be written

\begin{equation*} PA\cdot PB = PC\cdot PD \end{equation*}

If the intersection point is outside the circle.

The triangle \(APD\) is similar to triangle \(CPB.\) Hence \(AP/CP= PD/PB\) i.e. \(AP\cdot PB = PD\cdot CP\) which can be written

\begin{equation*} PA\cdot PB = PC\cdot PD. \end{equation*}

We have that \(\angle PTA = \angle TBA\) (alternate segment theorem) and \(\angle PTB = \angle TAP\) (angle sum of a triangle). Therefore triangle \(PTB\) is similar to triangle \(PAT.\) Hence \(PT/PA=PB/PT,\) which implies \(PT^2 = PA\cdot PB.\)

Subsection 3.15 Exercises