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Neokaal Study

Circles

Competency Practice

Tangent angle proofs, inradius of right triangles, chord-tangent angle relations, and cyclic quadrilateral proofs.

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  1. Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle TPQ = 2 \angle OPQ$.

  2. A circle with centre $O$ and radius $r$ is inscribed in a right triangle $\Delta ABC$, right-angled at $B$. If the lengths of the sides containing the right angle are $AB = 15\text{ cm}$ and $BC = 8\text{ cm}$, and hypotenuse $AC = 17\text{ cm}$, prove that $r = \frac{AB + BC - AC}{2}$ and hence find the value of $r$.

  3. From an external point T, tangent TA touches a circle with centre O. If $OT=12$ cm and $\angle OTA=30^\circ$, find TA.

    1. (A)$6\sqrt{3}$ cm
    2. (B)$6\sqrt{2}$ cm
    3. (C)$6$ cm
    4. (D)$12\sqrt{3}$ cm

  4. A circle has centre O. From an outside point P, the contact points of the two tangents are Q and R. Show that the four points Q, O, R, and P lie on one circle.

  5. Tangents PA and PB meet at P, and $\angle APB=70^\circ$. Find $\angle OAB$, where O is the centre.

    1. (A)110°
    2. (B)55°
    3. (C)65°
    4. (D)35°

  6. An outside point B is joined tangentially to a circle at C and D, with each tangent measuring 7 cm. The angle between them is $120^\circ$. Establish that the distance from B to the centre O is twice BC.

Neokaal Study

Answer Key

  1. $\angle TPQ = 2 \angle OPQ$

    1. Let $\angle PTQ = \theta$. By Theorem 10.2, $TP = TQ$, so $\triangle TPQ$ is isosceles with $\angle TPQ = \angle TQP$.
    2. In $\triangle TPQ$, $2\angle TPQ + \theta = 180^\circ \implies \angle TPQ = 90^\circ - \frac{\theta}{2}$.
    3. By Theorem 10.1, radius $OP \perp TP$, so $\angle OPT = 90^\circ$.
    4. Therefore $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\angle PTQ$.
    5. Multiplying both sides by $2$ gives $\angle PTQ = 2 \angle OPQ$ (Hence proved).

  2. Proved; $r = 3\text{ cm}$

    1. Let the incircle touch sides $AB, BC, AC$ at points $P, Q, R$ respectively.
    2. Since the radius to the point of contact is perpendicular to the tangent, $\angle OPB = \angle OQB = 90^\circ$. Along with $\angle B = 90^\circ$ and radii $OP = OQ = r$, quadrilateral $OPBQ$ is a square of side $r$, so $BP = BQ = r$.
    3. Lengths of tangents drawn from an external point to a circle are equal: $AP = AR = AB - r$ and $CQ = CR = BC - r$.
    4. Hypotenuse $AC = AR + CR = (AB - r) + (BC - r) = AB + BC - 2r \implies r = \frac{AB + BC - AC}{2}$ (Hence proved).
    5. Substituting the side lengths: $r = \frac{15 + 8 - 17}{2} = 3\text{ cm}$.

  3. $6\sqrt{3}$ cm

    1. Since OA is a radius to the point of contact, $OA\perp TA$.
    2. In right triangle OTA, $\cos 30^\circ=TA/12$.
    3. Therefore $TA=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}$ cm.

  4. Proved

    1. A radius is perpendicular to the tangent at its point of contact, so $\angle OQP=90^\circ$.
    2. Similarly, $\angle ORP=90^\circ$.
    3. These opposite angles sum to 180°, so QORP is a cyclic quadrilateral.

  5. 55°

    1. The equal tangents PA and PB make the configuration symmetric about OP.
    2. The angle between radii OA and OB is $180^\circ-70^\circ=110^\circ$.
    3. Triangle AOB is isosceles, so each base angle is $55^\circ$.

  6. Proved

    1. Triangles OCB and ODB are congruent, so BO bisects the 120° angle.
    2. Thus $\angle CBO=60^\circ$, while OC is perpendicular to BC.
    3. In right triangle BCO, $\cos60^\circ=BC/BO=1/2$, hence $BO=2BC$.