Answer: Since $\angle A = 70^\circ$ and $\angle B = 60^\circ$, the third angle is $\angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ$. Because all three angles ($70^\circ, 60^\circ, 50^\circ$) are less than $90^\circ$, $\Delta ABC$ is an acute-angled triangle. For an acute-angled triangle, the circumcentre lies inside the triangle.

Answer: Since $\angle A = 100^\circ$, which is greater than $90^\circ$, $\Delta ABC$ is an obtuse-angled triangle. For an obtuse-angled triangle, the circumcentre lies outside the triangle.

Answer: The least possible radius of a circle passing through two points $A$ and $B$ occurs when the line segment $AB$ itself is the diameter of the circle. Therefore, the least possible radius is half the length of the segment $AB$ ($AB/2$).

Answer: No, you cannot find a point $P$ such that $PA = PB = PC$.
The perpendicular bisector of $AB$ contains all points equidistant from $A$ and $B$, and the perpendicular bisector of $BC$ contains all points equidistant from $B$ and $C$. If $A, B$, and $C$ lie on the same straight line, the perpendicular bisectors of segments $AB$ and $BC$ will both be perpendicular to that same line. Since two lines perpendicular to the same line are parallel to each other, these bisectors will never intersect. Without an intersection point, there is no single point $P$ equidistant from all three.
Consequently, no circle can pass through three collinear points, and a straight line cannot cut a circle in three distinct points.

Answer: Let the circle have centre $C$ and let the chord be $AB$. Form $\Delta CAB$ by joining the centre $C$ to the endpoints of the chord $A$ and $B$. Both segments $CA$ and $CB$ connect the centre to points on the circle, so they are both radii of the circle. Since all radii of a given circle are equal in length, $CA = CB$. A triangle with two equal sides is an isosceles triangle.

Answer: Let the two triangles be $\Delta CAB$ and $\Delta CDE$, where $C$ is the centre of the circle, and $AB$ and $DE$ are two chords (the bases of the triangles).
1. $CA = CD$ (both are radii of the circle).
2. $CB = CE$ (both are radii of the circle).
3. $AB = DE$ (given that the bases are of equal length).
By the Side-Side-Side (SSS) congruence criterion, $\Delta CAB \cong \Delta CDE$.

Answer: Let the circle have centre $C$, chord $AB$, and let $CM$ be perpendicular to $AB$ (so $\angle CMA = \angle CMB = 90^\circ$). We want to show $AM = BM$.
Consider $\Delta CMA$ and $\Delta CMB$:
1. $CA = CB$ (radii of the same circle).
2. $CM = CM$ (common side).
3. $\angle CMA = \angle CMB = 90^\circ$ (given perpendicular).
By the Right-Angle Hypotenuse Side (RHS) congruence criterion, $\Delta CMA \cong \Delta CMB$. Therefore, by CPCTC, $AM = BM$, meaning the perpendicular bisects the chord.

Answer: Let $AD$ be the altitude from $A$ to $BC$. In the isosceles triangle $\Delta ABC$ ($AB = AC$), the altitude to the base $BC$ also bisects the base. Thus, $AD$ is the perpendicular bisector of chord $BC$. Since the perpendicular bisector of any chord of a circle must pass through the centre of that circle, the altitude $AD$ must pass through the centre.

Answer: Let the chords be $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$, and the radius be $r = 5\text{ cm}$. Let $O$ be the centre, and let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. A line through the centre perpendicular to parallel chords will pass through both their midpoints.
For chord $AB$: $AM = 6/2 = 3\text{ cm}$. In right $\Delta OMA$, $OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$.
For chord $CD$: $CN = 8/2 = 4\text{ cm}$. In right $\Delta ONC$, $ON = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$.
Since the chords are on opposite sides of the centre, the total distance between their midpoints is $OM + ON = 4 + 3 = \mathbf{7\text{ cm}}$.

Answer: The perpendiculars $CE$ and $CH$ from the centre $C$ to the chords $AB$ and $GF$ are given as equal ($CE = CH$). Theorem 7 states that chords of a circle that are equidistant from the centre have equal length. Therefore, $AB = GF$.

Answer: Draw radii $CA$ and $CG$. Both are equal to $r$.
In right $\Delta CEA$: $AE^2 = CA^2 - CE^2 = r^2 - CE^2$.
In right $\Delta CHG$: $GH^2 = CG^2 - CH^2 = r^2 - CH^2$.
Since we are given $CE = CH$, it follows that $AE^2 = GH^2$, so $AE = GH$.
Because a perpendicular from the centre bisects a chord, $AB = 2AE$ and $GF = 2GH$.
Therefore, $AB = GF$.

Answer: Let the radius be $r = 7\text{ cm}$ and the perpendicular distance to the chord be $d = 6\text{ cm}$. Let half the chord length be $x$.
Using Pythagoras' theorem: $x^2 + d^2 = r^2$
$x^2 + 6^2 = 7^2 \implies x^2 + 36 = 49 \implies x^2 = 13 \implies x = \sqrt{13}$.
The total chord length is $2x = \mathbf{2\sqrt{13}\text{ cm}}$ (or approx. $7.21\text{ cm}$).

Answer: The radius $r$, the perpendicular distance $d$ from the centre to the chord, and half the chord length ($x$) form a right-angled triangle. By the Baudhāyana-Pythagoras theorem, $x^2 + d^2 = r^2$, which gives $x^2 = r^2 - d^2$, so $x = \sqrt{r^2 - d^2}$. Since the perpendicular from the centre bisects the chord (Theorem 5), the total length of the chord is twice $x$, which is $2\sqrt{r^2 - d^2}$.

Answer: No, we cannot conclude that $CD = 2AB$.
The length of a chord based on its distance from the centre is given by $2\sqrt{r^2 - d^2}$. The relationship is non-linear due to the square and square root. While Theorem 8 states that a chord closer to the centre is longer, the length does not scale proportionally with the distance.

Answer: In $\Delta AOB$, $OA = OB = 12\text{ cm}$ (radii). This makes it an isosceles triangle, so $\angle OAB = \angle OBA$. Since $\angle AOB = 60^\circ$, the other two angles must also be $(180^\circ - 60^\circ) / 2 = 60^\circ$. Because all three angles are $60^\circ$, $\Delta AOB$ is an equilateral triangle. Therefore, the chord $AB$ is equal in length to the radii, so $AB = \mathbf{12\text{ cm}}$.

Answer: Let half the chord length be $x$. Using Pythagoras: $x = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$.
The total chord length is $2 \times 12 = \mathbf{24\text{ cm}}$.

Answer: By Theorem 9, the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
So, the angle on the circle = $70^\circ / 2 = \mathbf{35^\circ}$.

Answer: Radius $r = 26 / 2 = 13\text{ cm}$. Half the chord length $x = 24 / 2 = 12\text{ cm}$.
Let the distance from the centre be $d$. Using Pythagoras: $d = \sqrt{r^2 - x^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ cm}}$.

Answer: Half the chord length $x = \sqrt{r^2 - d^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$.
The total length of the chord is $2 \times 12 = \mathbf{24\text{ cm}}$.

Answer: $\angle ACB = \mathbf{90^\circ}$.
Reasoning: The angle subtended by an arc at the centre is double the angle subtended at any point on the circle. The arc $AB$ (being a semicircle formed by the diameter) subtends a straight angle of $180^\circ$ at the centre. Therefore, the angle $\angle ACB$ at the circumference must be half of $180^\circ$, which is $90^\circ$. This is also known as the corollary "the angle in a semicircle is a right angle".

Answer: By Theorem 11, the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.
$\angle C = 180^\circ - \angle A = 180^\circ - 75^\circ = \mathbf{105^\circ}$.
$\angle D = 180^\circ - \angle B = 180^\circ - 110^\circ = \mathbf{70^\circ}$.

Answer: Since it is a cyclic quadrilateral, opposite angles $\angle P$ and $\angle R$ sum to $180^\circ$.
$(2x + 10) + (3x - 20) = 180$
$5x - 10 = 180 \implies 5x = 190 \implies \mathbf{x = 38}$.
$\angle P = 2(38) + 10 = 76 + 10 = \mathbf{86^\circ}$.
$\angle R = 3(38) - 20 = 114 - 20 = \mathbf{94^\circ}$.
(Check: $86 + 94 = 180$).

Answer: Half the chord length $x = 16 / 2 = 8\text{ cm}$. Distance $d = 6\text{ cm}$.
Using Pythagoras: $r^2 = x^2 + d^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Radius $r = \sqrt{100} = \mathbf{10\text{ cm}}$.

Answer: This is a kite shape inscribed in a circle. The sides can form two right-angled triangles because $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. By the converse of Pythagoras, the angle between the sides of 5 and 12 is a right angle ($90^\circ$). Because the opposite angles must sum to $180^\circ$, they are both $90^\circ$.
The area is simply the sum of the areas of these two right triangles.
Area = $2 \times \left( \frac{1}{2} \times 5 \times 12 \right) = 5 \times 12 = \mathbf{60\text{ square units}}$.
*(Alternatively, use Brahmagupta's formula: $s = (5+5+12+12)/2 = 17$. Area = $\sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = 60$).*

Answer: Let $ABCD$ be a parallelogram inscribed in a circle. Because it is a parallelogram, opposite angles are equal ($\angle A = \angle C$). Because it is inscribed in a circle (cyclic), opposite angles sum to $180^\circ$ ($\angle A + \angle C = 180^\circ$).
Substituting gives $2\angle A = 180^\circ \implies \angle A = 90^\circ$. The same logic applies to $\angle B$ and $\angle D$. A parallelogram with all right angles is a rectangle. Therefore, only rectangles can be inscribed in a circle.

Answer: Let's look at vertex $D$. The interior angle is $\angle ADC$ and the exterior angle is $\angle CDE$. Since they form a linear pair on the straight line $CDE$, $\angle ADC + \angle CDE = 180^\circ$.
Because $ABCD$ is a cyclic quadrilateral, its opposite interior angles sum to $180^\circ$. Thus, $\angle ADC + \angle ABC = 180^\circ$.
Equating the two expressions gives: $\angle ADC + \angle CDE = \angle ADC + \angle ABC$. Subtracting $\angle ADC$ from both sides leaves $\angle CDE = \angle ABC$. The exterior angle is equal to the interior opposite angle.

Answer: Let the endpoints of any chord be $A$ and $B$, and let the centre of the circle be $C$ with radius $r$. Consider $\Delta CAB$. By the triangle inequality theorem, the length of any side of a triangle must be strictly less than the sum of the other two sides. So, $AB < CA + CB$.
Since $CA = r$ and $CB = r$, this means $AB < 2r$. Therefore, any chord that does not pass through the centre is strictly less than $2r$ (the diameter). If the chord passes through the centre, it *is* the diameter, and $A, C, B$ are collinear, so $AB = 2r$. Thus, no chord can be longer than the diameter.