In Fig. 10.13, and are two parallel tangents to a circle with centre and another tangent with point of contact intersecting at and at . Prove that .

Understand the Question and are parallel tangents touching the circle at and . The line segment joining the points of contact of two parallel tangents is a diameter, so is a straight line passing through centre ($ POQ = 180^$). Tangents drawn from an external point to a circle are equal in length ( and ). By joining , we can prove congruence of the pairs of triangles ($ OPA OCA OQB OCB POA = COA QOB = COB$. Summing all angles along the straight line yields . Step 1 · Prove Congruence of and Join . In and : (Radii of the same circle) (Tangents from an external point are equal) (Common side) By SSS congruence criterion: By CPCT: Step 2 · Prove Congruence of and In and : (Radii of the same circle) (Tangents from an external point are equal) (Common side) By SSS congruence criterion: By CPCT: Step 3 · Calculate Since , the line segment is a diameter passing through centre , making a straight line. Sum of angles on a straight line is : Using equations and : Since : Answer > Hence proved, . Common Mistakes Missing Construction: Forgetting to state the construction step (joining ), which is essential to form the triangles and . Unjustified Collinearity: Assuming is a straight line without explaining that the segment connecting the points of contact of two parallel tangents is a diameter. Congruence Criteria Confusion: Using tangent properties without clearly stating whether SSS or RHS criterion is being applied.

Circles | Exercise 10.2

Question 9

In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$ . Prove that $\angle AOB = 90^\circ$ .

Solution

Understand the Question

$XY$ and $X'Y'$ are parallel tangents touching the circle at $P$ and $Q$ . The line segment joining the points of contact of two parallel tangents is a diameter, so $POQ$ is a straight line passing through centre $O$ ( $\angle POQ = 180^\circ$ ).
Tangents drawn from an external point to a circle are equal in length ( $AP = AC$ and $BQ = BC$ ).
By joining $OC$ , we can prove congruence of the pairs of triangles ( $\triangle OPA \cong \triangle OCA$ and $\triangle OQB \cong \triangle OCB$ ) to show that $\angle POA = \angle COA$ and $\angle QOB = \angle COB$ .
Summing all angles along the straight line $POQ$ yields $\angle AOB = 90^\circ$ .

Step 1 · Prove Congruence of $\triangle OPA$ and $\triangle OCA$

Join $OC$ . Diagram 1

In $\triangle OPA$ and $\triangle OCA$ :

$OP = OC$ (Radii of the same circle)
$AP = AC$ (Tangents from an external point $A$ are equal)
$OA = OA$ (Common side)

By SSS congruence criterion: $\triangle OPA \cong \triangle OCA$

By CPCT: $\angle POA = \angle COA \quad \dots (1)$

Step 2 · Prove Congruence of $\triangle OQB$ and $\triangle OCB$

In $\triangle OQB$ and $\triangle OCB$ :

$OQ = OC$ (Radii of the same circle)
$BQ = BC$ (Tangents from an external point $B$ are equal)
$OB = OB$ (Common side)

By SSS congruence criterion: $\triangle OQB \cong \triangle OCB$

By CPCT: $\angle QOB = \angle COB \quad \dots (2)$

Step 3 · Calculate $\angle AOB$

Since $XY \parallel X'Y'$ , the line segment $PQ$ is a diameter passing through centre $O$ , making $POQ$ a straight line.

Sum of angles on a straight line is $180^\circ$ : $\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$

Using equations $(1)$ and $(2)$ :

\begin{aligned} \angle COA + \angle COA + \angle COB + \angle COB &= 180^\circ \\[0.6em] 2\angle COA + 2\angle COB &= 180^\circ \\[0.6em] 2(\angle COA + \angle COB) &= 180^\circ \\[0.6em] \angle COA + \angle COB &= 90^\circ \end{aligned}

Since $\angle COA + \angle COB = \angle AOB$ : $\angle AOB = 90^\circ$

Answer

Hence proved, $\angle AOB = 90^\circ$ .

Common Mistakes

Missing Construction: Forgetting to state the construction step (joining $OC$ ), which is essential to form the triangles $\triangle OCA$ and $\triangle OCB$ .
Unjustified Collinearity: Assuming $POQ$ is a straight line without explaining that the segment connecting the points of contact of two parallel tangents is a diameter.
Congruence Criteria Confusion: Using tangent properties without clearly stating whether SSS or RHS criterion is being applied.

Question 9

More questions in Exercise 10.2