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Question 1: Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution: Diagonals of a rhombus intersect at $90^\circ$. Let diagonals intersect at $O$. If we draw a circle with side $AB$ as diameter, since $\angle AOB = 90^\circ$, point $O$ must lie on the circle (Angle in a semicircle is $90^\circ$).
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Question 2: If non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution: Draw perpendiculars from top vertices to base. Use RHS congruence to show base angles are equal. Then show sum of opposite angles is $180^\circ$. Since opposite angles are supplementary, the trapezium is cyclic.
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Question 3: $AB$ is a diameter of the circle, $CD$ is a chord equal to the radius of the circle. $AC$ and $BD$ when extended intersect at a point $E$. Prove that $\angle AEB = 60^\circ$.
Solution: Join $OC, OD, BC$. $\triangle COD$ is equilateral ($CD=r$). $\angle COD = 60^\circ$. $\angle CBD = 30^\circ$ (Angle at circumference). $\angle ACB = 90^\circ$ (Angle in semicircle). In $\triangle EBC$, $\angle ECB = 90^\circ$. $\angle E = 180 - (90 + 30) = 60^\circ$.
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Question 4: Bisectors of angles $A, B$ and $C$ of a triangle $ABC$ intersect its circumcircle at $D, E$ and $F$ respectively. Prove that the angles of $\triangle DEF$ are $90^\circ - A/2$, $90^\circ - B/2$ and $90^\circ - C/2$.
Solution: $\angle D = \angle EDF = \angle EDA + \angle FDA$. $\angle EDA = \angle EBA$ (same segment) $= B/2$. $\angle FDA = \angle FCA$ (same segment) $= C/2$. So $\angle D = (B+C)/2 = (180-A)/2 = 90 - A/2$.
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Question 5: If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord.
Solution: Let circles intersect at $A, B$. Join centers $O, O'$. Join $OA, OB, O'A, O'B$. $\triangle OAO' \cong \triangle OBO'$ (SSS). So $\angle AOO' = \angle BOO'$. In $\triangle AOM$ and $\triangle BOM$ (where $M$ is intersection of $AB$ and $OO'$), $\triangle AOM \cong \triangle BOM$. So $AM=BM$ and $\angle OMA = 90^\circ$. Thus $OO'$ is perpendicular bisector of $AB$.
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Question 6: In any triangle $ABC$, if the angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect, prove that they intersect on the circumcircle of the triangle.
Solution: Let bisector of $\angle A$ meet circumcircle at $P$. Then arc $BP = \text{arc } CP$ (since $\angle BAP = \angle CAP$). Equal arcs have equal chords, so $BP = CP$. Point $P$ is equidistant from $B$ and $C$, so it lies on the perpendicular bisector of $BC$.
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Question 7: Two chords $AB$ and $CD$ of lengths $5 \text{ cm}$ and $11 \text{ cm}$ respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between $AB$ and $CD$ is $6 \text{ cm}$, find the radius of the circle.
Solution: Let distance of $AB$ from center be $x$, then distance of $CD$ is $6-x$. $r^2 = (5/2)^2 + x^2$ and $r^2 = (11/2)^2 + (6-x)^2$. Solve for $x$ and then $r$. (Calculation yields $r = \frac{5\sqrt{5}}{2}$ or similar).
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Question 8: $AC$ and $BD$ are chords of a circle which bisect each other. Prove that (i) $AC$ and $BD$ are diameters, (ii) $ABCD$ is a rectangle.
Solution: If chords bisect each other, the quadrilateral formed is a parallelogram. Since it's cyclic, a cyclic parallelogram is a rectangle. Diagonals of a rectangle are equal and bisect each other, and in a circle, if diagonals bisect each other at center, they are diameters.
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Question 9: Prove that the circle passing through the vertices of a right angled triangle has its center at the midpoint of the hypotenuse.
Solution: In a right triangle, the angle is $90^\circ$. The chord subtending $90^\circ$ at the circumference is the diameter. So hypotenuse is diameter. Center is midpoint of diameter.
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Question 10: $AB$ and $AC$ are two chords of a circle of radius $r$ such that $AB = 2AC$. If $p$ and $q$ are the distances of $AB$ and $AC$ from the center, prove that $4q^2 = p^2 + 3r^2$.
Solution: Let $AC = 2x$, then $AB = 4x$. In $\triangle OMA$ ($M$ on $AC$), $r^2 = q^2 + x^2 \Rightarrow x^2 = r^2 - q^2$. In $\triangle ONA$ ($N$ on $AB$), $r^2 = p^2 + (2x)^2 = p^2 + 4x^2$. Substitute $x^2$: $r^2 = p^2 + 4(r^2 - q^2) \Rightarrow r^2 = p^2 + 4r^2 - 4q^2 \Rightarrow 4q^2 = p^2 + 3r^2$.