This page provides comprehensive Chapter 10: Circles - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Circles. Practice for CBSE Board Exams.
Question 1: Prove that the lengths of tangents drawn from an external point to a circle are equal.
Solution: Step 1: Given a circle with centre O and a point P outside it. Two tangents PQ and PR are drawn. Step 2: Join OQ, OR and OP. $\angle OQP = \angle ORP = 90^\circ$ (Radius $\perp$ Tangent). Step 3: In right $\Delta OQP$ and $\Delta ORP$: OQ = OR (Radii) OP = OP (Common) $\Delta OQP \cong \Delta ORP$ (RHS congruence). Step 4: PQ = PR (CPCT). Hence proved.
Question 2: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution: Step 1: Let O be the centre. AB be the chord of larger circle touching smaller circle at P. Step 2: OP $\perp$ AB. In $\Delta OPA$, $OA^2 = OP^2 + AP^2$. Step 3: $5^2 = 3^2 + AP^2 \Rightarrow 25 = 9 + AP^2 \Rightarrow AP^2 = 16 \Rightarrow AP = 4$ cm. Step 4: Since OP bisects AB, Length AB $= 2 \times AP = 8$ cm.
Question 3: A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Solution: Step 1: Let points of contact be P, Q, R, S on sides AB, BC, CD, DA respectively. Step 2: Tangents from external point are equal: $AP=AS, BP=BQ, CR=CQ, DR=DS$. Step 3: Add LHS and RHS: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$. Step 4: $AB + CD = AD + BC$. Hence proved.
Question 4: Prove that the parallelogram circumscribing a circle is a rhombus.
Solution: Step 1: Let ABCD be the parallelogram. From Q3, we know $AB + CD = AD + BC$. Step 2: Since ABCD is a parallelogram, $AB = CD$ and $AD = BC$. Step 3: $2AB = 2AD \Rightarrow AB = AD$. Step 4: Since adjacent sides are equal, ABCD is a rhombus.
Question 5: In figure, 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: Step 1: Join OC. In $\Delta OPA$ and $\Delta OCA$, $OP=OC$ (radii), $OA=OA$, $AP=AC$ (tangents). Step 2: $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA$. Similarly $\angle QOB = \angle COB$. Step 3: POQ is a diameter line. $\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$. Step 4: $2\angle COA + 2\angle COB = 180^\circ \Rightarrow 2(\angle COA + \angle COB) = 180^\circ \Rightarrow \angle AOB = 90^\circ$.
Question 6: A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Solution: Step 1: Let tangents from A be $AE=AF=x$. $CD=CF=6, BD=BE=8$. Step 2: Sides: $a=14, b=x+6, c=x+8$. Semi-perimeter $s = 14+x$. Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(14+x)x(8)(6)} = \sqrt{48x(14+x)}$. Step 3: Area also $= \frac{1}{2}r(a+b+c) = 2(2x+28) = 4(x+14)$. Step 4: $16(x+14)^2 = 48x(x+14) \Rightarrow x+14 = 3x \Rightarrow 2x = 14 \Rightarrow x=7$. Answer: $AB=15$ cm, $AC=13$ cm.
Question 7: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution: Step 1: Let tangents be PA and PB touching at A and B. Centre O. Step 2: In quadrilateral OAPB, $\angle OAP = \angle OBP = 90^\circ$. Step 3: Sum of angles $= 360^\circ$. $\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ$. Step 4: $\angle APB + \angle AOB = 180^\circ$. Hence supplementary.
Question 8: 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 OT intersect PQ at R. PR = RQ = 4 cm. 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 $\Delta OPT$, $OT^2 = TP^2 + OP^2 \Rightarrow (y+3)^2 = x^2 + 25$. Step 3: Substitute $x^2$: $(y+3)^2 = y^2 + 16 + 25 \Rightarrow y^2 + 6y + 9 = y^2 + 41 \Rightarrow 6y = 32 \Rightarrow y = 16/3$. Step 4: $x^2 = (16/3)^2 + 16 = 256/9 + 144/9 = 400/9 \Rightarrow x = 20/3$. Answer: TP = 20/3 cm.
Question 9: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution: Step 1: Let XY be tangent at P to circle with centre O. Take point Q on XY other than P. Step 2: Q must lie outside the circle (otherwise XY would be secant). Step 3: Therefore, $OQ > OP$ (radius). This is true for every point on XY except P. Step 4: OP is the shortest distance from O to XY. Shortest distance is perpendicular. Hence $OP \perp XY$.
Question 10: From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that $\Delta APB$ is equilateral.
Solution: Step 1: Let radius be $r$. Then $OP = 2r$. In right $\Delta OAP$, $\sin(\angle APO) = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2}$. Step 2: $\angle APO = 30^\circ$. Since OP bisects $\angle APB$, $\angle APB = 2 \times 30^\circ = 60^\circ$. Step 3: In $\Delta APB$, $PA = PB$ (tangents), so $\angle PAB = \angle PBA$. Step 4: $2\angle PAB = 180^\circ - 60^\circ = 120^\circ \Rightarrow \angle PAB = 60^\circ$. Step 5: All angles are $60^\circ$, so $\Delta APB$ is equilateral.