NCERT Solutions: Class 9 Maths Chapter 9 - Circles

Let $AB$ and $CD$ be equal chords of congruent circles with centres $O$ and $O'$. Since the circles are congruent, $OA=OB=O'C=O'D$. Also $AB=CD$. Thus $\triangle AOB\cong\triangle CO'D$ by SSS congruence. Hence $\angle AOB=\angle CO'D$.

Let chords $AB$ and $CD$ of congruent circles with centres $O$ and $O'$ subtend equal angles, so $\angle AOB=\angle CO'D$. The radii are equal: $OA=O'C$ and $OB=O'D$. Therefore $\triangle AOB\cong\triangle CO'D$ by SAS congruence, and hence $AB=CD$.

Let the centres be $O$ and $O'$, with $OO'=4$ cm, and let $AB$ be the common chord. If $M$ is the midpoint of $AB$, then $OO'$ is perpendicular to $AB$. Put $OM=x$; then $O'M=4-x$. From the right triangles, $AM^2=5^2-x^2=3^2-(4-x)^2$. Hence $25-x^2=9-(4-x)^2$, giving $x=4$. Thus $AM^2=25-16=9$, so $AM=3$ cm and $AB=6$ cm.

Let perpendiculars from the centre $O$ to $AB$ and $CD$ meet them at $M$ and $N$. Equal chords are equidistant from the centre, so $OM=ON$, and perpendiculars from the centre bisect chords, so $AM=BM$ and $CN=DN$. In right triangles $OMP$ and $ONP$, $OP$ is common and $OM=ON$, so $\triangle OMP\cong\triangle ONP$ by RHS. Hence $PM=PN$. Combining equal half-chords with $PM=PN$ gives the corresponding chord segments equal: $AP=CP$ and $BP=DP$.

Let equal chords $AB$ and $CD$ intersect at $P$, and let $O$ be the centre. Draw $OM\perp AB$ and $ON\perp CD$. Equal chords are equidistant from the centre, so $OM=ON$. In right triangles $OMP$ and $ONP$, $OP$ is common; hence $\triangle OMP\cong\triangle ONP$ by RHS. Therefore $\angle OPM=\angle OPN$, so $OP$ makes equal angles with the two chords.

Let the line meet the outer circle at $A,D$ and the inner circle at $B,C$. Draw $OM$ perpendicular to the line. Since a perpendicular from the centre to a chord bisects the chord, $M$ is the midpoint of both $AD$ and $BC$. Hence $AM=MD$ and $BM=MC$. Therefore $AB=AM-BM$ and $CD=DM-CM$, so $AB=CD$.

Equal chords $RS$ and $SM$ subtend equal central angles. For chord $RS=6$ m in a circle of radius $5$ m, half the chord is $3$ m, so if $\angle ROS=\theta$, then $\sin\dfrac{\theta}{2}=\dfrac{3}{5}$ and $\cos\dfrac{\theta}{2}=\dfrac{4}{5}$. Chord $RM$ subtends angle $2\theta$, so $RM=2(5)\sin\theta=10\left(2\cdot\dfrac35\cdot\dfrac45\right)=9.6$ m.

Three equally spaced points on a circle form an equilateral triangle. Each side is a chord subtending $120^\circ$ at the centre. Therefore each string length is $2R\sin60^\circ=2(20)\cdot\dfrac{\sqrt3}{2}=20\sqrt3$ m.

The central angle subtending arc $ABC$ is $\angle AOC=\angle AOB+\angle BOC=60^\circ+30^\circ=90^\circ$. The angle at the circumference standing on the same arc is half the central angle. Hence $\angle ADC=\dfrac12\times90^\circ=45^\circ$.

If chord $AB$ equals the radius, then $OA=OB=AB$, so $\triangle AOB$ is equilateral and the minor central angle $\angle AOB=60^\circ$. At a point on the major arc, the chord subtends half of $60^\circ$, i.e. $30^\circ$. At a point on the minor arc, it subtends half of the major arc angle $360^\circ-60^\circ=300^\circ$, i.e. $150^\circ$.

$\angle PQR=100^\circ$ subtends the major arc $PR$, whose central angle is $200^\circ$. Hence the minor central angle $\angle POR=360^\circ-200^\circ=160^\circ$. Since $OP=OR$, triangle $OPR$ is isosceles, so $\angle OPR=\angle ORP=\dfrac{180^\circ-160^\circ}{2}=10^\circ$.

In $\triangle ABC$, $\angle BAC=180^\circ-69^\circ-31^\circ=80^\circ$. Angles in the same segment of a circle are equal, so $\angle BDC=\angle BAC=80^\circ$.

Since $AC$ and $BD$ intersect at $E$, $\angle CED=180^\circ-\angle BEC=50^\circ$. In $\triangle CED$, $\angle CDE=180^\circ-50^\circ-20^\circ=110^\circ$. Thus $\angle CDB=110^\circ$. The angles $\angle BAC$ and $\angle BDC$ stand on chord $BC$ from opposite segments, so they are supplementary. Hence $\angle BAC=180^\circ-110^\circ=70^\circ$.

Angles in the same segment give $\angle DAC=\angle DBC=70^\circ$. Thus $\angle DAB=\angle DAC+\angle CAB=70^\circ+30^\circ=100^\circ$. Opposite angles of a cyclic quadrilateral are supplementary, so $\angle BCD=180^\circ-100^\circ=80^\circ$. If $AB=BC$, then $\triangle ABC$ is isosceles and $\angle ACB=\angle BAC=30^\circ$. Since $E$ lies on $AC$, $\angle ECD=\angle ACD=\angle BCD-\angle BCA=80^\circ-30^\circ=50^\circ$.

Let $ABCD$ be cyclic and let both diagonals $AC$ and $BD$ be diameters. An angle in a semicircle is a right angle. Since $BD$ is a diameter, $\angle BAD=\angle BCD=90^\circ$. Since $AC$ is a diameter, $\angle ABC=\angle ADC=90^\circ$. All four angles are right angles, so $ABCD$ is a rectangle.

Let $ABCD$ be a trapezium with $AB\parallel CD$ and $AD=BC$. In an isosceles trapezium, base angles are equal, so $\angle A=\angle B$ and $\angle C=\angle D$. Also, because $AB\parallel CD$, consecutive interior angles are supplementary: $\angle A+\angle D=180^\circ$. Since $\angle C=\angle D$, we get $\angle A+\angle C=180^\circ$. A quadrilateral with a pair of opposite angles supplementary is cyclic. Hence $ABCD$ is cyclic.

In the circle through $A,B,C,P$, angles standing on chord $AP$ are equal, so $\angle ACP=\angle ABP$. Since $A,B,D$ are collinear and $P,B,Q$ are collinear, $\angle ABP=\angle DBQ$. In the circle through $D,B,C,Q$, angles standing on chord $DQ$ are equal, so $\angle DBQ=\angle QCD$. Therefore $\angle ACP=\angle QCD$.

Let $ABC$ be a triangle and let circles be drawn with $AB$ and $AC$ as diameters. Let the circles meet again at $D$. Since $AB$ is a diameter, $\angle ADB=90^\circ$. Since $AC$ is a diameter, $\angle ADC=90^\circ$. Thus both $DB$ and $DC$ are perpendicular to $AD$ at $D$, so $B,D,C$ are collinear. Hence $D$ lies on the third side $BC$.

Since $ABC$ and $ADC$ are right triangles with common hypotenuse $AC$, the points $A,B,C,D$ lie on the same circle with $AC$ as diameter. In this circle, $\angle CAD$ and $\angle CBD$ stand on the same chord $CD$. Therefore, angles in the same segment are equal, so $\angle CAD=\angle CBD$.

In a parallelogram, opposite angles are equal. In a cyclic quadrilateral, opposite angles are supplementary. If $ABCD$ is both cyclic and a parallelogram, then $\angle A=\angle C$ and $\angle A+\angle C=180^\circ$. Hence $2\angle A=180^\circ$, so $\angle A=90^\circ$. Similarly all angles are right angles. Therefore the parallelogram is a rectangle.