For a circle to pass through A and B, AB is a chord. The longest chord of a circle is its diameter, so the smallest circle through A and B is obtained when AB itself is the diameter. Hence the least radius is half of AB.
The least possible radius is $\dfrac{AB}{2}$.
Let AB be a chord of a circle with centre O. Join OA and OB. Since A and B lie on the circle, $OA=OB$ because both are radii. Therefore $\triangle AOB$ has two equal sides and is isosceles.
Let the two chords be AB and DE in the same circle with centre O, and suppose $AB=DE$. Join OA, OB, OD and OE. Since all four are radii, $OA=OB=OD=OE$. Also $AB=DE$. Thus $\triangle AOB$ and $\triangle DOE$ have three corresponding sides equal, so they are congruent by SSS.
Let C be the centre and let CM be perpendicular to chord AB. Join CA and CB. Since CA and CB are radii, $CA=CB$. Also $CM$ is common and $\angle CMA=\angle CMB=90°$. Therefore $\triangle CMA \cong \triangle CMB$ by RHS congruence, so $AM=BM$. Hence the perpendicular from the centre bisects the chord.
In an isosceles triangle, the altitude from the vertex A to the base BC also bisects BC. A line perpendicular to chord BC through its midpoint passes through the centre of the circle. Hence the altitude from A to BC must pass through the centre.
Since $AB=AC$, A lies on the perpendicular bisector of BC. The centre of the circle is also equidistant from B and C, so it lies on the perpendicular bisector of BC. The altitude from A to BC in an isosceles triangle is this same perpendicular bisector. Therefore the altitude from A to BC passes through the centre.
The perpendicular from the centre to a chord bisects the chord. For the chord of length 6 cm, half-chord $=3$ cm, so its distance from the centre is $\sqrt{5^2-3^2}=\sqrt{16}=4$ cm. For the chord of length 8 cm, half-chord $=4$ cm, so its distance from the centre is $\sqrt{5^2-4^2}=\sqrt{9}=3$ cm. Since the chords are on opposite sides of the centre, the distance between their midpoints is $4+3=7$ cm.
The distance between the midpoints is $7$ cm.
Let AB and DE be equal chords. Let their midpoints be M and N. Then $AM=DN$ because the chords are equal and each is bisected by the perpendicular from the centre. In right triangles OMA and OND, the hypotenuse is the radius and one leg is the half-chord. By the Baudhāyana-Pythagoras theorem, $OM^2=r^2-AM^2$ and $ON^2=r^2-DN^2$. Since $AM=DN$, $OM=ON$.
Theorem 6 says that equal chords of a circle are equidistant from the centre. If two equal chords have length $2a$ in a circle of radius r, the perpendicular from the centre bisects each chord. For each chord, the distance d from the centre satisfies $d^2+a^2=r^2$, so $d=\sqrt{r^2-a^2}$. Since r and a are the same for both chords, their distances from the centre are equal.
Since C is the centre, $CA=CF$ are radii. Also $CE=CH$ and both triangles $\triangle CEA$ and $\triangle CHF$ are right triangles. By RHS congruence, $\triangle CEA \cong \triangle CHF$, so $AE=HF$. The perpendicular from the centre bisects each chord, so $AB=2AE$ and $GF=2HF$. Hence $AB=GF$.
Let the radius be r. Since $CE=CH$, the half-length of chord AB is $AE=\sqrt{r^2-CE^2}$ and the half-length of chord GF is $HF=\sqrt{r^2-CH^2}$. Because $CE=CH$, we get $AE=HF$. Therefore $AB=2AE=2HF=GF$.
The perpendicular from the centre bisects the chord. Half the chord is $\sqrt{7^2-6^2}=\sqrt{49-36}=\sqrt{13}$ cm. Therefore the full chord is $2\sqrt{13}$ cm.
The chord length is $2\sqrt{13}$ cm.
The perpendicular from the centre to a chord bisects the chord. If half the chord is x, the radius r, distance d and half-chord x form a right triangle. So $x^2+d^2=r^2$, hence $x=\sqrt{r^2-d^2}$. Therefore the full chord length is $2x=2\sqrt{r^2-d^2}$.
For a concrete counterexample, take $r=5$ cm. If AB is at distance $4$ cm, then $AB=2\sqrt{25-16}=6$ cm. If CD is at distance $2$ cm, then $CD=2\sqrt{25-4}=2\sqrt{21}$ cm, which is not $2AB=12$ cm. Hence the conclusion is false.
No. Chord length is not directly proportional to distance from the centre. If the radius is r and the distance from the centre is d, the chord length is $2\sqrt{r^2-d^2}$, not a linear expression in d.
Since $OA=OB=12$ cm and $\angle AOB=60°$, triangle AOB has two equal sides with included angle $60°$. The base angles are also $60°$, so $\triangle AOB$ is equilateral. Therefore $AB=12$ cm.
$AB=12$ cm.
Part (i) follows from the theorem that the same chord AB subtends equal angles at points on the same arc. For (ii) and (iii), equality of angles alone is not enough without the correct position condition. The converse theorem used for concyclicity requires the points to lie on the same side of AB.
(i) No. Angles in the same segment of a circle are equal.
(ii) Not always; equal angles can also occur in special cases such as when AB is a diameter, where points on opposite arcs subtend right angles.
(iii) Not necessarily. If X and Y are on the same side of AB and subtend equal angles at AB, then A, B, X and Y are concyclic; without that condition, the conclusion need not follow.
In Fig. 5.26, the quadrilateral is cyclic and the angle opposite x is $100°$. Opposite angles of a cyclic quadrilateral are supplementary, so $x+100°=180°$. Hence $x=80°$.
$x=80°$.
Half the chord is $\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm. Therefore the chord length is $2\times 12=24$ cm.
The chord length is $24$ cm.
The angle subtended by an arc at a point on the circle is half the angle subtended by the same arc at the centre. Therefore the required angle is $\dfrac{70°}{2}=35°$.
$35°$.
The radius is $13$ cm and half the chord is $12$ cm. If the distance from the centre to the chord is d, then $d^2+12^2=13^2$. So $d^2=169-144=25$ and $d=5$ cm.
The distance is $5$ cm.
Half the chord is $\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12$ cm. Therefore the full chord length is $24$ cm.
The chord length is $24$ cm.
Let AB be a chord and O be the centre. Since A and B are on the circle, $OA=OB$. A point that is equidistant from A and B lies on the perpendicular bisector of AB. Therefore O lies on the perpendicular bisector of AB, so the perpendicular bisector of the chord passes through the centre.
The diameter AB subtends a straight angle of $180°$ at the centre. The angle subtended by the same arc at a point on the circle is half the central angle. Therefore $\angle ACB=\dfrac{180°}{2}=90°$.
$\angle ACB=90°$.
Opposite angles of a cyclic quadrilateral are supplementary. So $\angle C=180°-75°=105°$ and $\angle D=180°-110°=70°$.
$\angle C=105°$ and $\angle D=70°$.
In a cyclic quadrilateral, opposite angles are supplementary. So $(2x+10)+(3x-20)=180$. This gives $5x-10=180$, $5x=190$, and $x=38$. Then $\angle P=2(38)+10=86°$ and $\angle R=3(38)-20=94°$.
$x=38$, $\angle P=86°$ and $\angle R=94°$.
Half the chord is $8$ cm. Radius, half-chord and perpendicular distance form a right triangle, so $r^2=8^2+6^2=64+36=100$. Hence $r=10$ cm.
The radius is $10$ cm.
For a cyclic quadrilateral, Brahmagupta's formula gives area $=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is the semiperimeter. Here $s=\dfrac{5+5+12+12}{2}=17$. Area $=\sqrt{(17-5)(17-5)(17-12)(17-12)}=\sqrt{12\times12\times5\times5}=60$ square units.
The area is $60$ square units.
If two numbers have sum S and product T, they are roots of $u^2-Su+T=0$. Since the segment-pairs $(AP,PB)$ and $(CP,PD)$ have the same sum and product, they are the same two roots. Hence the corresponding segments are equal, possibly after matching the longer segment with the longer segment and the shorter with the shorter.
Let equal chords AB and CD intersect at P. Then $AP+PB=CP+PD$ because $AB=CD$. Also, by the intersecting chords theorem, $AP\cdot PB=CP\cdot PD$. Two positive segment-pairs with the same sum and same product are the same pair of numbers. Therefore the two segments of one chord are equal to the corresponding two segments of the other chord.