Basic properties of circles  (10.510.6)
Understand the properties of tangents to a circle and angles in the alternate segments

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

ExamplesIn 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)

Theorythe straight line perpendicular to a radius of a circle at its external extremity is a tangent to the circle
(converse of tangent \(\perp\) radius)

ExamplesIn 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 \(\angle\)s, isos. )
\(OB = OA\) (radii)
\(\begin{array}{1}\angle OBA = \angle OAB\\ = 20^\circ \end{array}\) (base \(\angle\)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)

Theorythe perpendicular to a tangent at its point of contact passes through the centre of the circle

ExamplesIn 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. 
SolutionsIn \(\Delta APT\),
\(\begin{array}{1}\angle ATP + \angle PAT + \angle APT = 180^\circ \\\angle ATP + 50^\circ + 40^\circ = 180^\circ \\\angle ATP = 90^\circ \end{array}\) ( 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

Theoryif two tangents are drawn to a circle from an external point, then: TANGENT PROP

ExamplesIn 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

Theoryif a straight line is tangent to a circle, then the tangentchord angle is equal to the angle in the alternate segment

ExamplesIn 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)

Theoryif 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

ExamplesIn 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 \(\angle\)s, isos. )
\(\angle CBA = \angle DCB\) (alt. \(\angle\)s, CD // AB)
\(\angle CAB = \angle DCB\)
\(\therefore\) CD is a tangent to the circle at C. (converse of \(\angle\)s in alt. segment)
Use the basic properties of circles to perform simple geometric proofs

Theory

ExamplesIn the figure, AB is the tangent to circle CDE at E. If AB // CD, prove that \(CE = DE\).

Solutions\(\angle ECD = \angle DEB\) (\(\angle\) in alt. segment)
\(\angle EDC = \angle DEB\) (alt. \(\angle\)s, AB // CD)
\(\therefore \angle ECD = \angle EDC\)
\(\therefore CE = DE\) (sides opp. eq. \(\angle\)s)