Class 10 Circles HOTS Worksheet

Question 1: In the given figure, $PQ$ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length $TP$.

Solution:
Step 1: Let $TR$ intersect $PQ$ at $R$. $TR \perp PQ$ and bisects it. $PR = 4$. In $\Delta ORP$, $OR = \sqrt{5^2-4^2} = 3$.
Step 2: Let $TP = x, TR = y$. In $\Delta PRT$, $x^2 = y^2 + 16$. In $\Delta OPT$, $TP^2 + OP^2 = OT^2 \Rightarrow x^2 + 25 = (y+3)^2$.
Step 3: Substitute $x^2$: $y^2 + 16 + 25 = y^2 + 6y + 9 \Rightarrow 41 = 6y + 9 \Rightarrow 6y = 32 \Rightarrow y = 16/3$.
Step 4: $x^2 = (16/3)^2 + 16 = 256/9 + 144/9 = 400/9 \Rightarrow x = 20/3$ cm.
Question 2: Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$.

Solution:
Step 1: Let $\angle PTQ = \theta$. Since $TP = TQ$, $\Delta TPQ$ is isosceles. $\angle TPQ = \angle TQP = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2}$.
Step 2: Radius is perpendicular to tangent, so $\angle OPT = 90^\circ$.
Step 3: $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - (90^\circ - \frac{\theta}{2}) = \frac{\theta}{2}$.
Step 4: Thus $\theta = 2\angle OPQ$, i.e., $\angle PTQ = 2\angle OPQ$.
Question 3: A circle touches the side $BC$ of $\Delta ABC$ at $P$ and touches $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ = \frac{1}{2}(\text{Perimeter of } \Delta ABC)$.

Solution:
Step 1: Tangents from external point are equal. $AQ = AR$, $BQ = BP$, $CP = CR$.
Step 2: Perimeter $= AB + BC + AC = AB + (BP + PC) + AC$.
Step 3: Substitute $BP=BQ, PC=CR$: Perimeter $= (AB + BQ) + (AC + CR) = AQ + AR$.
Step 4: Since $AQ=AR$, Perimeter $= 2AQ \Rightarrow AQ = \frac{1}{2}(\text{Perimeter})$.
Question 4: Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

Solution:
Step 1: Let $XY$ and $X'Y'$ be parallel tangents touching at $P$ and $Q$. Let tangent $AB$ touch at $C$, intersecting $XY$ at $A$ and $X'Y'$ at $B$.
Step 2: Join $OC, OA, OB$. $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA$.
Step 3: $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB$.
Step 4: $\angle POQ = 180^\circ$. $2\angle COA + 2\angle COB = 180^\circ \Rightarrow \angle COA + \angle COB = 90^\circ \Rightarrow \angle AOB = 90^\circ$.
Question 5: In the given figure, a circle is inscribed in a quadrilateral $ABCD$ in which $\angle B = 90^\circ$. If $AD = 23$ cm, $AB = 29$ cm and $DS = 5$ cm, find the radius $r$ of the circle.

Solution:
Step 1: Let points of contact be $P, Q, R, S$ on $AB, BC, CD, DA$. $DS = DR = 5$.
Step 2: $AR = AD - DR = 23 - 5 = 18$. $AQ = AR = 18$ (Tangents from A).
Step 3: $BQ = AB - AQ = 29 - 18 = 11$.
Step 4: Since $\angle B = 90^\circ$, $OQBP$ is a square with side $r$. So $r = BQ = 11$ cm.
Question 6: $AB$ is a diameter and $AC$ is a chord of a circle with centre $O$ such that $\angle BAC = 30^\circ$. The tangent at $C$ intersects extended $AB$ at a point $D$. Prove that $BC = BD$.

Solution:
Step 1: Join $OC$. $OA=OC \Rightarrow \angle OCA = \angle OAC = 30^\circ$. $\angle COB = 60^\circ$ (Exterior angle).
Step 2: In $\Delta OCD$, $\angle OCD = 90^\circ$. $\angle D = 180 - (90+60) = 30^\circ$.
Step 3: In $\Delta ABC$, $\angle ACB = 90^\circ$ (angle in semicircle), $\angle ABC = 60^\circ$.
Step 4: $\angle CBD = 180 - 60 = 120^\circ$. In $\Delta BCD$, $\angle BCD = 180 - (120+30) = 30^\circ$.
Step 5: Since $\angle BCD = \angle D = 30^\circ$, $BC = BD$.
Question 7: If an isosceles triangle $ABC$, in which $AB = AC = 6$ cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Solution:
Step 1: Let $AD$ be altitude. $O$ is centre on $AD$. $OA=9$. Let $OD=x$.
Step 2: In $\Delta OBD$, $BD^2 = 9^2 - x^2$. In $\Delta ABD$, $BD^2 = 6^2 - (9-x)^2$.
Step 3: $81 - x^2 = 36 - (81 - 18x + x^2) \Rightarrow 81 - x^2 = 36 - 81 + 18x - x^2 \Rightarrow 126 = 18x \Rightarrow x = 7$.
Step 4: $BD = \sqrt{81 - 49} = \sqrt{32} = 4\sqrt{2}$. $BC = 8\sqrt{2}$. Height $AD = 9-7=2$.
Step 5: Area $= \frac{1}{2} \times 8\sqrt{2} \times 2 = 8\sqrt{2}$ cm$^2$.
Question 8: A chord $PQ$ of a circle is parallel to the tangent drawn at a point $R$ of the circle. Prove that $R$ bisects the arc $PRQ$.

Solution:
Step 1: Let tangent be $XY$ at $R$. $PQ || XY$.
Step 2: $\angle PRX = \angle PQR$ (Alternate segment theorem).
Step 3: $\angle PRX = \angle RPQ$ (Alternate interior angles).
Step 4: So $\angle PQR = \angle RPQ$. Sides opposite equal angles are equal: $PR = RQ$.
Step 5: Equal chords cut equal arcs. Arc $PR$ = Arc $RQ$.
Question 9: The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle.

Solution:
Step 1: Let sides be $a, b, c$. Segments of $a$ are 6, 8. So $a = 14$. Let segments of $b$ be 8, $x$ and $c$ be 6, $x$.
Step 2: $s = 14+x$. Area $= \sqrt{48x(14+x)}$. Also Area $= rs = 4(14+x)$.
Step 3: $16(14+x)^2 = 48x(14+x) \Rightarrow 14+x = 3x \Rightarrow 2x = 14 \Rightarrow x = 7$.
Step 4: Sides: $6+7=13$ cm, $8+7=15$ cm.
Question 10: $PQ$ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length of $TP$.

Solution:
Step 1: $PR = 8$. $OR = \sqrt{10^2 - 8^2} = 6$.
Step 2: Let $TP = x, TR = y$. $x^2 = y^2 + 64$. $x^2 + 100 = (y+6)^2$.
Step 3: $y^2 + 64 + 100 = y^2 + 12y + 36 \Rightarrow 164 = 12y + 36 \Rightarrow 12y = 128 \Rightarrow y = 32/3$.
Step 4: $x^2 = (32/3)^2 + 64 = 1024/9 + 576/9 = 1600/9 \Rightarrow x = 40/3$ cm.

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Chapter 10: Circles

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