Basic properties of circles - (10.5-10.6)

Understand the properties of tangents to a circle and angles in the alternate segments

  • Graph
    a tangent to a circle is perpendicular to the radius through the point of contact



  • Examples
    In the figure, O is the centre of the circle. TA and TB are tangents
    to the circle at A and B respectively. Find x.
  • Solutions
    \(\angle OAT = 90^\circ \) (tangent \perp radius)
    \(\angle OBT = 90^\circ \) (tangent \perp radius)
    \(\begin{array}{1}\angle AOB + \angle OAT + \angle OBT + x = (4 - 2) \times 180^\circ \\124^\circ + 90^\circ + 90^\circ + x = 360{^\circ {}}\\x = {\underline{\underline {56^\circ }} {}}\end{array}\) (\angle sum of polygon)
  • Theory
    the straight line perpendicular to a radius of a circle at its external extremity is a tangent to the circle
    (converse of tangent \perp radius)
  • Examples
    In the figure, O is the centre of the circle. TA is the tangent to the circle at A. B is a
    point on the circumference such that \(TA = TB\). It is given that \(\angle OAB = 20^\circ \).
    Prove that TB is the tangent to the circle at B.
  • Solutions
    \(\angle OAT = 90^\circ \) (tangent  radius)
    \(\begin{array}{1}\angle BAT + 20^\circ = 90^\circ \\\angle BAT = 70^\circ \end{array}\)
    \(TB = TA\) (given)
    \(\begin{array}{1}\angle ABT = \angle BAT\\ = 70^\circ \end{array}\) (base s, isos. )
    \(OB = OA\) (radii)
    \(\begin{array}{1}\angle OBA = \angle OAB\\ = 20^\circ \end{array}\) (base s, isos. )
    \(\begin{array}{1}\angle OBT = \angle OBA + \angle ABT\\ = 20^\circ + 70^\circ \\ = 90^\circ \end{array}\)
    \(\therefore\) B is the tangent to the circle at B. (converse of tangent  radius)
  • Graph
  • Theory
    the perpendicular to a tangent at its point of contact passes through the centre of the circle
  • Examples
    In the figure, A is a point on the circumference. PQ is the tangent to the
    circle at T. \(\angle APT = 50^\circ \) and \(\angle PAT = 40^\circ \). Prove that AT is a diameter.
  • Solutions
    In \(\Delta APT\),
    \(\begin{array}{1}\angle ATP + \angle PAT + \angle APT = 180^\circ \\\angle ATP + 50^\circ + 40^\circ = 180^\circ \\\angle ATP = 90^\circ \end{array}\) ( sum of )
    \(AT \bot PQ\)
    \(\therefore\) AT passes through the centre of the circle.
    (line passing through the point of contact  tangent passes through centre)
    \(\therefore\) AT is a diameter.
  • Graph
  • Theory
    if two tangents are drawn to a circle from an external point, then: TANGENT PROP



  • Examples
    In the figure, the inscribed circle of ABC touches AC, AB and BC at D, E and F respectively.
    It is given that \(AB = 32{\rm{ cm}}\), \(AC = 26{\rm{ cm}}\), \(AE = x{\rm{ cm}}\) and \(BC = (3x - 12){\rm{ cm}}\).
    Find the length of BC.
  • Solutions
    \(BE = (32 - x){\rm{ cm}}\)
    \(BF = BE\) (tangents from ext. pt.)
    \( = (32 - x){\rm{ cm}}\)
    \(AD = AE\) (tangents from ext. pt.)
    \( = x{\rm{ cm}}\)
    \(CD = (26 - x){\rm{ cm}}\)
    \(CF = CD\) (tangents from ext. pt.)
    \( = (26 - x){\rm{ cm}}\)

    \(\begin{array}{1}BC = BF + CF\\3x - 12 = 32 - x + 26 - x\\3x - 12 = 58 - 2x\\5x = 70\\x = 14\end{array}\)

    \( \begin{array}{1}\therefore BC = (3 \times 14 - 12){\rm{ cm}}\\ = \underline{\underline {30{\rm{ cm}}}} \end{array}\)
  • Graph
  • Theory
    if a straight line is tangent to a circle, then the tangent-chord angle is equal to the angle in the alternate segment
  • Examples
    In the figure, AB is the tangent to the circle at T. C and D are points on
    the circumference. \(\angle DCT = 7x - 26^\circ \) and \(\angle BTD = 5x\). Find x.
  • Solutions
    \(\begin{array}{1}\angle DCT = \angle BTD\\7x - 26^\circ = 5{x^{}}\\2x = 26{^\circ {}}\\x = {\underline{\underline {13^\circ }} {}}\end{array}\) ( in alt. segment)
  • Graph
  • Theory
    if a straight line passes through an end point of a chord of a circle that the angle it makes with the chord is equal to the angle in the alternate segment, then the straight line touches the circle
  • Examples
    In the figure, \(CA = CB\) and AB // CD. Prove that CD is the tangent to the circle at C.
  • Solutions
    \(CA = CB\) (given)
    \(\angle CAB = \angle CBA\) (base s, isos. )
    \(\angle CBA = \angle DCB\) (alt. s, CD // AB)
    \(\angle CAB = \angle DCB\)
    \(\therefore\) CD is a tangent to the circle at C. (converse of  in alt. segment)

Use the basic properties of circles to perform simple geometric proofs

  • Theory
  • Examples
    In the figure, AB is the tangent to circle CDE at E. If AB // CD, prove that \(CE = DE\).
  • Solutions
    \(\angle ECD = \angle DEB\) ( in alt. segment)
    \(\angle EDC = \angle DEB\) (alt. s, AB // CD)
    \(\therefore \angle ECD = \angle EDC\)
    \(\therefore CE = DE\) (sides opp. eq. s)