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Question 1: If the length of a chord of a circle is $16 \text{ cm}$ and is at a distance of $15 \text{ cm}$ from the center of the circle, find the radius of the circle.
Solution: Perpendicular from center bisects the chord. Half chord $= 8 \text{ cm}$. Distance $= 15 \text{ cm}$. By Pythagoras theorem, $r^2 = 8^2 + 15^2 = 64 + 225 = 289$. So $r = 17 \text{ cm}$.
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Question 2: In a cyclic quadrilateral $ABCD$, if $\angle A = (2x + 4)^\circ$ and $\angle C = (4x - 64)^\circ$, find the value of $x$ and the angles.
Solution: Sum of opposite angles is $180^\circ$. $\angle A + \angle C = 180^\circ \Rightarrow 2x + 4 + 4x - 64 = 180 \Rightarrow 6x - 60 = 180 \Rightarrow 6x = 240 \Rightarrow x = 40$. $\angle A = 84^\circ, \angle C = 96^\circ$.
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Question 3: Prove that equal chords of a circle subtend equal angles at the center.
Solution: Let chords $AB = CD$. In $\triangle AOB$ and $\triangle COD$: $OA=OC$ (radii), $OB=OD$ (radii), $AB=CD$ (given). By SSS congruence, $\triangle AOB \cong \triangle COD$. Thus $\angle AOB = \angle COD$ (CPCT).
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Question 4: $ABCD$ is a cyclic quadrilateral in which $AC$ and $BD$ are its diagonals. If $\angle DBC = 55^\circ$ and $\angle BAC = 45^\circ$, find $\angle BCD$.
Solution: $\angle DAC = \angle DBC = 55^\circ$ (Angles in same segment). $\angle DAB = \angle DAC + \angle BAC = 55^\circ + 45^\circ = 100^\circ$. Since $ABCD$ is cyclic, $\angle DAB + \angle BCD = 180^\circ \Rightarrow \angle BCD = 180^\circ - 100^\circ = 80^\circ$.
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Question 5: Two circles intersect at two points $A$ and $B$. $AD$ and $AC$ are diameters to the two circles. Prove that $B$ lies on the line segment $DC$.
Solution: Join $AB$. Since $AD$ is diameter, $\angle ABD = 90^\circ$ (Angle in semicircle). Since $AC$ is diameter, $\angle ABC = 90^\circ$. $\angle ABD + \angle ABC = 90^\circ + 90^\circ = 180^\circ$. Thus $DBC$ is a straight line, so $B$ lies on $DC$.
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Question 6: If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution: Since diagonals are diameters, they intersect at the center. Also, angle in a semicircle is $90^\circ$. All four angles of the quadrilateral are subtended by diameters, so each angle is $90^\circ$. Hence it is a rectangle.
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Question 7: Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.
Solution: Let $ABCD$ be a quadrilateral. Bisectors of $\angle A$ and $\angle B$ meet at $P$. In $\triangle APB$, $\angle P = 180 - \frac{1}{2}(A+B)$. Similarly for $\triangle CRD$, $\angle R = 180 - \frac{1}{2}(C+D)$. Sum of opposite angles $\angle P + \angle R = 360 - \frac{1}{2}(A+B+C+D) = 360 - \frac{1}{2}(360) = 180^\circ$. Hence cyclic.
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Question 8: Two congruent circles intersect each other at points $A$ and $B$. Through $A$ any line segment $PAQ$ is drawn so that $P, Q$ lie on the two circles. Prove that $BP = BQ$.
Solution: Since circles are congruent, equal arcs subtend equal angles. Arc $AB$ in both circles is equal (common chord). So $\angle APB = \angle AQB$ (Angles in same segment of congruent circles). In $\triangle PBQ$, angles opposite to sides are equal, so $BP = BQ$.
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Question 9: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution: Let chords $AB, CD$ intersect at $P$. Draw $OM \perp AB, ON \perp CD$. In $\triangle OMP$ and $\triangle ONP$: $OM=ON$ (equal chords equidistant), $OP=OP$, $\angle M = \angle N = 90^\circ$. So $\triangle OMP \cong \triangle ONP$. Thus $MP=NP$. Since $AM=CN$ (half of equal chords), $AP=CP$ and $BP=DP$.
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Question 10: Prove that a cyclic parallelogram is a rectangle.
Solution: Let $ABCD$ be a cyclic parallelogram. $\angle A + \angle C = 180^\circ$ (Cyclic). $\angle A = \angle C$ (Opposite angles of parallelogram). So $2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$. A parallelogram with one angle $90^\circ$ is a rectangle.