Circle properties and the tangent-chord theorem questions with answers

Questions

1. Question 1Angles from the tangent-chord theorem

   $P$, $Q$ and $R$ lie on the circle. The tangent to the circle at $P$ meets $RQ$ produced at $X$. $\angle XPQ = 58^\circ$ and $\angle PXQ = 22^\circ$. Find $\angle QPR$, giving a reason for each step.

   Show the worked solution

   1. Step 1: The angle between the tangent at P and the chord PQ$\angle PRQ = \angle XPQ$ (alt. segment theorem)$\angle PRQ = 58^\circ$
   2. Step 2: Find ∠PQR using triangle PQX$\mathstrut \angle PQR = \angle XPQ + \angle PXQ$ (ext. ∠ of △)$\angle PQR = 80^\circ$
   3. Step 3: The angles of triangle PQR add up to 180°$\mathstrut \angle QPR = 180^\circ - 58^\circ - 80^\circ$ (∠ sum of △)$\angle QPR = 42^\circ$
2. Question 2Angles from the tangent-chord theorem

   $TX$ and $TY$ are tangents to the circle at $X$ and $Y$, and $Z$ lies on the major arc $XY$. $\angle XTY = 64^\circ$. Find $\angle XZY$, giving a reason for each step.

   Show the worked solution

   1. Step 1: Tangents from T are equal, so triangle TXY is isosceles$TX = TY$ (tangents from ext. point)$\angle TXY = \angle TYX$ (base ∠s of isos. △)$\mathstrut \angle TXY = \dfrac{180^\circ - 64^\circ}{2}$$\angle TXY = 58^\circ$
   2. Step 2: The angle between the tangent at X and the chord XY$\angle XZY = \angle TXY$ (alt. segment theorem)$\angle XZY = 58^\circ$
3. Question 3Tangent and secant: PT² = PA × PB

   $PT$ is the tangent to the circle at $T$, and the straight line $PAB$ cuts the circle at $A$ and $B$. $PA = 8$ cm and $PT = 12$ cm. Prove that $\triangle PTA$ is similar to $\triangle PBT$, and hence that $PT^{2} = PA \times PB$. Find $AB$.

   Show the worked solution

   1. Step 1: Two pairs of equal angles, each with its reason$\angle TPA = \angle BPT$ (common ∠)$\angle PTA = \angle PBT$ (alt. segment theorem)
   2. Step 2: Name the similarity test$\triangle PTA$ is similar to $\triangle PBT$ (AA).
   3. Step 3: Matching sides are in the same ratio: write each side of △PTA over its match$\dfrac{PT}{PB} = \dfrac{PA}{PT}$$\dfrac{12}{PB} = \dfrac{8}{12}$$PB = 18$
   4. Step 4: AB is what is left of PB$\mathstrut AB = 18 - 8$$AB = 10$ cm
4. Question 4Tangent and secant: PT² = PA × PB

   $PT$ is the tangent to the circle at $T$, and the straight line $PAB$ cuts the circle at $A$ and $B$. $PA = 4$ cm and $AB = 12$ cm. Prove that $\triangle PTA$ is similar to $\triangle PBT$, and hence that $PT^{2} = PA \times PB$. Find $PT$.

   Show the worked solution

   1. Step 1: Two pairs of equal angles, each with its reason$\angle TPA = \angle BPT$ (common ∠)$\angle PTA = \angle PBT$ (alt. segment theorem)
   2. Step 2: Name the similarity test$\triangle PTA$ is similar to $\triangle PBT$ (AA).
   3. Step 3: PB is the whole line: PA + AB$\mathstrut PB = 4 + 12$$PB = 16$
   4. Step 4: Matching sides are in the same ratio: write each side of △PTA over its match$\dfrac{PT}{PB} = \dfrac{PA}{PT}$$\dfrac{PT}{16} = \dfrac{4}{PT}$
   5. Step 5: Multiply out: PT² = PA × PB$PT^{2} = 64$$PT = 8$ cm
5. Question 5Intersecting chords and secants

   The straight lines $TKL$ and $TMN$ cut the circle at $K$ and $L$, and at $M$ and $N$. $TK = 4$ cm, $KL = 8$ cm and $TM = 6$ cm. Prove that $\triangle TKM$ is similar to $\triangle TNL$, and hence that $TK \times TL = TM \times TN$. Find $MN$.

   Show the worked solution

   1. Step 1: Two pairs of equal angles, each with its reason$\angle KTM = \angle NTL$ (common ∠)$\angle TKM = \angle TNL$ (ext. ∠ of cyclic quad.)
   2. Step 2: Name the similarity test$\triangle TKM$ is similar to $\triangle TNL$ (AA).
   3. Step 3: TL is the whole line: TK + KL$\mathstrut TL = 4 + 8$$TL = 12$
   4. Step 4: Matching sides are in the same ratio: write each side of △TNL over its match$\dfrac{TN}{TK} = \dfrac{TL}{TM}$, so $TK \times TL = TM \times TN$$\dfrac{TN}{4} = \dfrac{12}{6}$
   5. Step 5: Work out TN$TN = 8$
   6. Step 6: MN is what is left of TN$\mathstrut MN = 8 - 6$$MN = 2$ cm
6. Question 6Intersecting chords and secants

   The chords $PR$ and $QS$ of a circle meet at $E$. $PE = 6$ cm, $ER = 4$ cm and $QE = 8$ cm. Prove that $\triangle PEQ$ is similar to $\triangle SER$, and hence that $PE \times ER = QE \times ES$. Find $ES$.

   Show the worked solution

   1. Step 1: Two pairs of equal angles, each with its reason$\angle QPE = \angle RSE$ (∠s in same segment)$\angle PEQ = \angle SER$ (vert. opp. ∠s)
   2. Step 2: Name the similarity test$\triangle PEQ$ is similar to $\triangle SER$ (AA).
   3. Step 3: Matching sides are in the same ratio: write each side of △SER over its match$\dfrac{ES}{EP} = \dfrac{ER}{EQ}$, so $PE \times ER = QE \times ES$$\dfrac{ES}{6} = \dfrac{4}{8}$
   4. Step 4: Work out ES$ES = 3$ cm