Class 9 Mathematics Chapter 5: I'm Up and Down, and Round and Round

A circle is made up of every point that lies one fixed distance from a chosen point, all of them in a single plane. The two quantities it depends on are that given point, called the centre, and that fixed distance, called the radius.

A chord is any line segment joining two points on the circle, and a diameter does exactly that, passing through the centre on its way. So a diameter is a particular chord. Its length is twice the radius, and it is the longest chord the circle has.

The longest chord is the diameter, because two points on a circle can be no further apart than the distance straight across it. The diameter is twice the radius, so the longest chord is 30 cm.

An arc is a part of the circle itself, a curved line running between two points on it. A segment is a region of the interior, bounded by a chord on one side and the arc joining the ends of that chord on the other. So an arc is a line and has length, while a segment is an area.

There is exactly one circle passing through three points that are not in a straight line. The condition cannot be dropped because three points lying in a straight line have no point of the plane equidistant from all three, so no centre exists and no circle can be drawn through them.

Fold the paper in half so that the two edges meet exactly, and crease it. Open it out and fold again in a different direction, creasing once more. The point where the two creases cross is the centre. The method works because a fold that matches the two halves exactly is a line of symmetry of the circle, and every line of symmetry of a circle passes through its centre, so each crease is a diameter. Two distinct lines meet at exactly one point, so their crossing must be the centre.

By the Baudhāyana–Pythagoras theorem the hypotenuse is $\sqrt{36 + 64} = \sqrt{100} = 10$ cm. In a right-angled triangle the circumcentre sits exactly halfway along the hypotenuse, so the centre is that midpoint, and the circumradius is half the hypotenuse, namely 5 cm. Placing the right angle at the origin with the legs along the axes, the vertices are $(0,0)$, $(8,0)$ and $(0,6)$, and the centre is $(4, 3)$, which is indeed 5 units from each.

The centre of any such circle would have to be the same distance from all three points. The points equidistant from the first two lie on one line, and the points equidistant from the second and third lie on another. When the three points are not in a straight line those two lines are not parallel, so they cross at exactly one point, and that point is the only candidate for the centre. Once the centre is fixed, its distance to the points fixes the radius, so the circle is completely determined and there is exactly one.

For an acute-angled triangle the circumcentre lies inside the triangle. For a right-angled triangle it lies exactly at the midpoint of the hypotenuse. For an obtuse-angled triangle it lies outside the triangle altogether. The special consequence is in the right-angled case: since the centre is the midpoint of the hypotenuse, the circumradius is exactly half the hypotenuse.

A circle has rotational symmetry about its centre through every angle, and a line of symmetry through the centre in every direction. So the circle looks exactly the same whichever way it is turned or reflected. Any chord can therefore be carried to any other chord of the same length by such a movement, which cannot change any property of the figure. Nothing about which direction a chord points can matter, leaving only its length and its distance from the centre to distinguish one chord from another.

(i) A circle is the set of all points in a plane at a fixed distance from a given point, the centre; that fixed distance is the radius. A chord is a line segment joining two points on the circle. A diameter is a chord that passes through the centre.

(ii) A chord joins two points of the circle, and no two points of a circle can be further apart than the distance straight across it, which is $2r$. That distance is achieved exactly when the chord passes through the centre, so the diameter is the longest chord.

(iii) An arc is part of the circle itself, a curved line between two points on it. A segment is a region of the interior bounded by a chord and the arc joining its ends. A sector is a region bounded by two radii and the arc between them.

(iv) A circle has infinitely many lines of symmetry — every line through the centre is one. It also has rotational symmetry about its centre through any angle at all, not merely through certain special angles.

(v) Fold the sheet in half so the edges meet exactly, then unfold and fold again in a different direction. Each fold matches the two halves, so each crease is a line of symmetry and therefore a diameter passing through the centre. Two distinct lines cross at exactly one point, so the point where the creases meet is the centre.

(i) Infinitely many. The centre may be placed anywhere, and the radius then chosen so that the circle reaches the point.

(ii) Still infinitely many, but with less freedom: the centre must be equidistant from the two points, which restricts it to one line, and any point of that line will serve.

(iii) There is exactly one circle passing through three points that are not in a straight line.

(iv) The centre would have to be equidistant from all three. The points equidistant from each pair form lines, and when the three original points are collinear those lines come out parallel, so they never meet. With no point equidistant from all three there is no possible centre, and so no circle.

(v) Once three of the four points are chosen, and provided they are not in a line, Theorem 1 fixes the circle completely — there is no freedom left. The fourth point must then happen to fall exactly on that already-determined circle, which a randomly chosen point will not generally do.

(i) The centre must be the same distance from all three markers, since that common distance is the radius of the circle through them.

(ii) Find the set of points equidistant from the first two markers, which is a line, and then the set equidistant from another pair, which is a second line. Because the markers are not in a straight line these two lines are not parallel, so they cross at exactly one point, and that point is the centre.

(iii) The circle is the circumcircle of the triangle formed by the three markers, and its centre is the circumcentre.

(iv) In a right-angled triangle the circumcentre sits exactly halfway along the hypotenuse, so the radius is half the longest side: 13 m.

(v) For an obtuse-angled triangle the circumcentre lies outside the triangle altogether, so the simple halving no longer applies and the radius would have to be found from the construction in (ii) rather than read off the longest side.

(i) The longest chord is the diameter, which is twice the radius, so 24 cm.

(ii) No. Every chord is at most as long as the diameter, which here is 24 cm, and 25 cm exceeds that. Two points of the circle simply cannot be 25 cm apart.

(iii) Infinitely many. Every line through the centre meets the circle in two points and so gives a diameter, and there are infinitely many lines through a point.

(iv) Joining the two ends of the chord to the centre forms an angle there, and that angle is the one the chord subtends at the centre. The word subtend means stands on.

(v) A circle is unchanged by any rotation about its centre and by reflection in any line through the centre. Given two chords of the same length, one can be carried exactly onto the other by such a movement, which changes nothing about the figure. So the two are indistinguishable as far as the circle is concerned, and only length and distance from the centre can tell chords apart.

If the radius is $r$, the chord has length $\ell$ and its distance from the centre is $d$, then $\left(\dfrac{\ell}{2}\right)^2 + d^2 = r^2$. It is the theorem of the right triangle applied to the triangle made by the radius, half the chord and the perpendicular from the centre.

Half the chord squared is $13^2 - 5^2 = 169 - 25 = 144$, so half the chord is 12 cm. The chord itself is therefore $2 \times 12 = 24$ cm.

Theorem 2 says equal chords of a circle subtend equal angles at the centre. Theorem 3 says chords subtending equal angles at the centre are equal. Each is the converse of the other: one begins with equal chords and ends with equal angles, while the other begins with equal angles and ends with equal chords.

Join the centre to both ends of the chord. The two triangles formed have equal radii as their hypotenuses, a common side along the perpendicular, and a right angle at the foot, so they are congruent. The remaining sides, which are the two parts of the chord, are therefore equal, and the foot is the midpoint.

The 24 cm chord is nearer. In the relation the squares of half the chord and of the distance add to the fixed value $r^2$, so a larger half-chord leaves a smaller distance. This is Theorem 8: the longer of two unequal chords is the nearer to the centre.

Half the chord is 8 cm, so the distance satisfies $d^2 = 10^2 - 8^2 = 100 - 64 = 36$, giving $d = 6$ cm. By Theorem 7, any chord of the same circle lying at that same distance from the centre must have the same length, so any such chord also measures 16 cm.

Half of 48 is 24, and $25^2 - 24^2 = 625 - 576 = 49$, so the distance is 7 cm. The 14 cm chord is shorter, and by Theorem 8 the shorter of two unequal chords is the further from the centre, so it must lie more than 7 cm away. Indeed half of it is 7, and $\sqrt{625 - 49} = 24$ cm.

Let $AB$ and $CD$ be equal chords of a circle with centre $O$. Join $OA$, $OB$, $OC$ and $OD$. In the triangles $OAB$ and $OCD$ we have $OA = OC$ and $OB = OD$, since all four are radii of the same circle, and $AB = CD$ by assumption. The three sides of one triangle therefore equal the three sides of the other, so the triangles are congruent. Corresponding angles of congruent triangles are equal, so the angle $AOB$ equals the angle $COD$, which is what was to be proved.

Half the chord is 12 cm, so the radius satisfies $r^2 = 12^2 + 16^2 = 144 + 256 = 400$, giving $r = 20$ cm. The longest chord of any circle is its diameter, which measures twice the radius, so it is 40 cm.

In a given circle the radius is a fixed number, so the relation $\left(\tfrac{\ell}{2}\right)^2 + d^2 = r^2$ ties the chord and the distance together completely: fixing either one fixes the other. Theorem 6 puts in two equal chord lengths and gets the same distance out, and Theorem 7 puts in two equal distances and gets the same length out. Nothing else about a chord appears in the relation, so nothing else can affect the answer.

(i) If the radius is $r$, the chord is $\ell$ and the distance from the centre is $d$, then $\left(\dfrac{\ell}{2}\right)^2 + d^2 = r^2$.

(ii) Half the chord enters the relation. The perpendicular from the centre bisects the chord, so the right-angled triangle formed has the radius as hypotenuse and half the chord as one of its legs.

(iii) Half the chord squared is $400 - 144 = 256$, so half the chord is 16 cm and the chord itself is 32 cm.

(iv) Of two unequal chords of a circle, the longer one is nearer to the centre.

(v) A diameter passes through the centre, so its distance from the centre is zero, the smallest value possible. It is also the longest chord the circle has. Longest and nearest together are exactly what Theorem 8 predicts, so the diameter is the extreme case of the rule rather than an exception to it.

(i) The line joining the centre of a circle to the midpoint of a chord is perpendicular to that chord.

(ii) The perpendicular drawn from the centre of a circle to a chord bisects that chord.

(iii) Theorem 4 is given a midpoint and produces a right angle. Theorem 5 is given a right angle and produces a midpoint. The two facts are the same pair of properties taken in opposite orders.

(iv) Theorem 6: two chords of equal length stand equally far from the centre. Theorem 7: two chords standing equally far from the centre have equal length.

(v) A proof has to move from what is given to what is wanted. Quoting the converse means assuming the very thing that was to be shown, so the chain of reasoning does not connect even if the number at the end happens to be correct. The marks for reasoning are awarded for that chain, not for the answer.

(i) Half the rod squared is $625 - 225 = 400$, so half is 20 cm and the rod measures 40 cm.

(ii) It lies 15 cm from the centre as well. By Theorem 6, chords of equal length in one circle are all the same distance from the centre, and the direction the rod points in makes no difference.

(iii) Further from the centre. The third rod is shorter, and by Theorem 8 the shorter of two unequal chords is the further one.

(iv) Half of 14 is 7, so the distance squared is $625 - 49 = 576$ and the distance is 24 cm, which is indeed more than 15 cm.

(v) The longest is 50 cm, the diameter of the pipe, and it would pass straight through the centre at distance zero.

(i) Let $OM$ be perpendicular to the chord $AB$, with $M$ on $AB$. In the triangles $OAM$ and $OBM$, $OA = OB$ as radii, $OM$ is common, and the angles at $M$ are both right angles. Two right-angled triangles with equal hypotenuses and one equal leg are congruent, so $AM = MB$ and $M$ is the midpoint.

(ii) It would need the three equal sides instead: $OA = OB$ as radii, $AM = MB$ because $M$ is given as the midpoint, and $OM$ common. Congruence then makes the two angles at $M$ equal, and since they lie on a straight line and add to $180°$, each is $90°$.

(iii) The relation would require $\left(\tfrac{\ell}{2}\right)^2 = 81 - 100$, a negative number, which no square can be. A chord can never lie further from the centre than the radius, so 10 cm is impossible in a circle of radius 9 cm.

(iv) It gets shorter. The two squares in the relation add to the fixed value $r^2$, so as the distance grows the half-chord must shrink.

(v) A circle is carried exactly onto itself by any rotation about its centre, and such a rotation carries any chord onto another chord of the same length and the same distance from the centre. Since the whole figure is unchanged, no property can depend on the direction, leaving only length and distance to matter.

Theorem 9 says the angle an arc makes at the centre is twice the angle that same arc makes at any point left over on the circle. So an arc subtending $140°$ at the centre subtends half of that, namely $70°$, at any point on the major arc.

The angle at the centre is double the angle at any point on the remaining part of the circle, so it is $2 \times 54° = 108°$.

A diameter passes through the centre, so its two ends and the centre lie on a straight line and the angle it subtends at the centre is $180°$. By Theorem 9 the angle it subtends at any other point of the circle is half of that, which is $90°$.

Opposite angles of a cyclic quadrilateral add to $180°$, so the opposite angle is $180° - 112° = 68°$. The rule ties each pair of opposite angles together on its own, so the remaining pair has no bearing on this calculation.

Points are concyclic when they all lie on one and the same circle. Theorem 11 establishes it: if a segment subtends equal angles at two points lying on the same side of the line containing it, then those two points together with the ends of the segment are concyclic.

Let $AB$ be a chord of a circle with centre $O$, and let $P$ and $Q$ be two points in the same segment. By Theorem 9 the angle at the centre standing on the arc $AB$ is double the angle at $P$, so the angle $APB$ is half the angle $AOB$. The same theorem applied to $Q$ gives the angle $AQB$ as half the very same angle $AOB$, since there is only one such angle at the centre. Two quantities each equal to half of one fixed quantity must be equal, so the angle $APB$ equals the angle $AQB$.

Since $AB$ is a diameter, the angle at $C$ is a right angle, so $\angle ACB = 90°$. The three angles of a triangle add to $180°$, so $\angle BAC = 180° - 90° - 38° = 52°$. The two remaining angles are therefore $90°$ at $C$ and $52°$ at $A$.

Take a cyclic quadrilateral $ABCD$. The angles at $A$ and at $C$ stand on the two arcs $BCD$ and $BAD$, which between them make up the whole circle. By Theorem 9 each angle is half the angle its arc subtends at the centre, and the two angles at the centre together make a full turn of $360°$. Halving, the two angles at $A$ and $C$ add to $180°$. The same argument applied to the other pair of arcs gives the same result for $B$ and $D$.

For a cyclic quadrilateral each pair of opposite angles must add to $180°$. Taking the angles in order, the first and third are opposite, and $95° + 85° = 180°$. The second and fourth are also opposite, and $70° + 110° = 180°$. Both pairs satisfy the condition, so by the converse of Theorem 12 the quadrilateral is cyclic.

Opposite angles of a parallelogram are equal to each other. If the parallelogram is also cyclic then the same opposite angles must add to $180°$. Two equal numbers adding to $180°$ must each be $90°$, so every angle of the parallelogram is a right angle and it is a rectangle. Conversely a rectangle is always cyclic, since $90° + 90° = 180°$ for both pairs.

(i) The angle an arc makes at the centre is twice the angle that same arc makes at any point left over on the circle.

(ii) The angle at the centre is the larger. Going from the circle in to the centre you double, and coming back out you halve.

(iii) Half of $130°$ is $65°$.

(iv) Theorem 10 says angles in the same segment are equal. Each such angle is half the one angle at the centre standing on the same arc, and two quantities each half of one fixed quantity must agree.

(v) When the chord is a diameter its ends and the centre lie in a straight line, so it subtends $180°$ at the centre. Halving by Theorem 9 gives $90°$ at any other point of the circle.

(i) A quadrilateral whose four vertices all lie on one circle.

(ii) The sum of either pair of opposite angles of a cyclic quadrilateral is $180°$.

(iii) $180° - 118° = 62°$.

(iv) If one pair of opposite angles of a quadrilateral adds to $180°$, the quadrilateral is cyclic. It is the standard way of proving that four given points lie on one circle.

(v) The opposite angles of a parallelogram are equal, and being cyclic they must also add to $180°$. Two equal numbers with that sum are each $90°$, so all four angles are right angles and the figure is a rectangle.

(i) $90°$. The line $AB$ passes through the centre and so is a diameter, and the angle standing on a diameter from any other point of the circle is a right angle.

(ii) The angles of the triangle add to $180°$, so the angle at $A$ is $180° - 90° - 27° = 63°$.

(iii) $180°$. All four points lie on the circle, so $ABCD$ is a cyclic quadrilateral and its opposite angles are supplementary.

(iv) A circle is the set of points at exactly the fixed radius from the centre. A point inside the rim is nearer the centre than that, so it fails the defining condition and cannot lie on the circle.

(v) They must lie on the same side of $AB$ as each other, in the same segment. Angles in the same segment are equal, so from any two such points the segment $AB$ subtends the same angle; here, $AB$ being a diameter, that common angle is $90°$.

(i) The four points $A$, $B$, $P$ and $Q$ lie on one circle, that is, they are concyclic. This is Theorem 11, the converse of the equal-angles-in-one-segment result.

(ii) From opposite sides of $AB$ the two angles are not equal but supplementary, since they would then be opposite angles of a cyclic quadrilateral. Equality is what characterises a single segment, so the same-side condition cannot be dropped.

(iii) No. The two given angles add to $195°$ rather than $180°$, and the converse of Theorem 12 requires a sum of exactly $180°$ for the quadrilateral to be cyclic.

(iv) The angle at any point of the arc is half the angle the chord subtends at the centre, and that angle at the centre does not depend in any way on where the point is. Half of an unchanging quantity is itself unchanging.

(v) Either show that a segment subtends equal angles at two points on the same side of it, which is Theorem 11; or show that one pair of opposite angles of the quadrilateral formed adds to $180°$, which is the converse of Theorem 12.

A circle is made up of every point that lies one fixed distance from a chosen point, all of them in a single plane. The two quantities it depends on are that given point, called the centre, and that fixed distance, called the radius.

A chord is any line segment joining two points on the circle, and a diameter does exactly that, passing through the centre on its way. So a diameter is a particular chord. Its length is twice the radius, and it is the longest chord the circle has.

The longest chord is the diameter, because two points on a circle can be no further apart than the distance straight across it. The diameter is twice the radius, so the longest chord is 30 cm.

If the radius is $r$, the chord has length $\ell$ and its distance from the centre is $d$, then $\left(\dfrac{\ell}{2}\right)^2 + d^2 = r^2$. It is the theorem of the right triangle applied to the triangle made by the radius, half the chord and the perpendicular from the centre.

Half the chord squared is $13^2 - 5^2 = 169 - 25 = 144$, so half the chord is 12 cm. The chord itself is therefore $2 \times 12 = 24$ cm.

Theorem 2 says equal chords of a circle subtend equal angles at the centre. Theorem 3 says chords subtending equal angles at the centre are equal. Each is the converse of the other: one begins with equal chords and ends with equal angles, while the other begins with equal angles and ends with equal chords.

Theorem 9 says the angle an arc makes at the centre is twice the angle that same arc makes at any point left over on the circle. So an arc subtending $140°$ at the centre subtends half of that, namely $70°$, at any point on the major arc.

The angle at the centre is double the angle at any point on the remaining part of the circle, so it is $2 \times 54° = 108°$.

A diameter passes through the centre, so its two ends and the centre lie on a straight line and the angle it subtends at the centre is $180°$. By Theorem 9 the angle it subtends at any other point of the circle is half of that, which is $90°$.