### Basic properties of circles - (10.5-10.6)

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

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

• 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.
• $$\angle OAT = 90^\circ$$ (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)
• the straight line perpendicular to a radius of a circle at its external extremity is a tangent to the circle
• 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.
• $$\angle OAT = 90^\circ$$ (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)
• the perpendicular to a tangent at its point of contact passes through the centre of the circle
• 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.
• 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.
• if two tangents are drawn to a circle from an external point, then: TANGENT PROP

• 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.
• $$BE = (32 - x){\rm{ cm}}$$
$$BF = BE$$ (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.)
$$= (26 - x){\rm{ cm}}$$

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

$$\begin{array}{1}\therefore BC = (3 \times 14 - 12){\rm{ cm}}\\ = \underline{\underline {30{\rm{ cm}}}} \end{array}$$
• if a straight line is tangent to a circle, then the tangent-chord angle is equal to the angle in the alternate segment
• 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.
• $$\begin{array}{1}\angle DCT = \angle BTD\\7x - 26^\circ = 5{x^{}}\\2x = 26{^\circ {}}\\x = {\underline{\underline {13^\circ }} {}}\end{array}$$ ( in alt. segment)
• 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
• In the figure, $$CA = CB$$ and AB // CD. Prove that CD is the tangent to the circle at C.
• $$CA = CB$$ (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

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