Ganit Prakash - Class-X - Theorems related to circle
Let us work out 3.2 Geometry
📘 Exercise 3.2 Solutions (Q1 - Q17)
Chapter 3The length of a radius of a circle with its centre O is 5 cm. and the length of its chord AB is 8 cm. Let us write by calculating, the distance of the chord AB from the centre O.
Explanation:
Let the circle have centre $O$. The radius $OB$ = 5 cm and the chord $AB$ = 8 cm.
Let $OD$ be the perpendicular distance from the centre $O$ to the chord $AB$.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
$$ \begin{array}{l} \text{Therefore, } DB = \frac{AB}{2} = \frac{8}{2} = 4 \text{ cm.} \\ \text{In the right-angled } \triangle ODB\text{, by Pythagoras theorem:} \\ OD^2 + DB^2 = OB^2 \\ OD^2 + 4^2 = 5^2 \\ OD^2 + 16 = 25 \\ OD^2 = 25 - 16 = 9 \\ OD = \sqrt{9} = 3 \text{ cm.} \end{array} $$Answer: The distance of the chord $AB$ from the centre $O$ is 3 cm.
The length of a diameter of a circle with its centre at O is 26 cm. The distance of the chord PQ from the point O is 5 cm. Let us write by calculating, the length of the chord PQ.
Explanation:
The diameter of the circle is 26 cm.
$$ \begin{array}{l} \text{Radius } (OQ) = \frac{26}{2} = 13 \text{ cm.} \end{array} $$Let $OD$ be the perpendicular distance from the centre $O$ to the chord $PQ$. We are given $OD$ = 5 cm.
$$ \begin{array}{l} \text{In the right-angled } \triangle ODQ\text{, by Pythagoras theorem:} \\ DQ^2 + OD^2 = OQ^2 \\ DQ^2 + 5^2 = 13^2 \\ DQ^2 + 25 = 169 \\ DQ^2 = 169 - 25 = 144 \\ DQ = \sqrt{144} = 12 \text{ cm.} \end{array} $$Since the perpendicular from the centre bisects the chord, $PQ = 2 \times DQ$.
$$ \begin{array}{l} PQ = 2 \times 12 = 24 \text{ cm.} \end{array} $$Answer: The length of the chord $PQ$ is 24 cm.
The length of a chord PQ of a circle with its centre O is 4 cm. and the distance of PQ from the point O is 2.1 cm. Let us write by calculating, the length of its diameter.
Explanation:
The length of the chord $PQ$ = 4 cm. The perpendicular distance from centre $O$ is $OD$ = 2.1 cm.
$$ \begin{array}{l} \text{Since } OD \perp PQ\text{, it bisects the chord. So, } DQ = \frac{4}{2} = 2 \text{ cm.} \\ \text{In the right-angled } \triangle ODQ\text{, by Pythagoras theorem:} \\ OQ^2 = OD^2 + DQ^2 \\ OQ^2 = (2.1)^2 + 2^2 \\ OQ^2 = 4.41 + 4 \\ OQ^2 = 8.41 \\ OQ = \sqrt{8.41} = 2.9 \text{ cm.} \end{array} $$The radius of the circle is 2.9 cm.
$$ \begin{array}{l} \text{Diameter} = 2 \times \text{Radius} = 2 \times 2.9 = 5.8 \text{ cm.} \end{array} $$Answer: The length of its diameter is 5.8 cm.
The lengths of two chords of a circle with its centre O are 6 cm. and 8 cm. If the distance of smaller chord from centre is 4 cm., then let us write by calculating, the distance of other chord from the centre.
Explanation:
Let the smaller chord be $AB$ = 6 cm and its perpendicular distance from the centre be $OD$ = 4 cm.
$$ \begin{array}{l} \text{Half of the chord } DB = \frac{6}{2} = 3 \text{ cm.} \\ \text{In right-angled } \triangle ODB\text{, radius } OB^2 = OD^2 + DB^2 \\ OB^2 = 4^2 + 3^2 = 16 + 9 = 25 \\ OB = \sqrt{25} = 5 \text{ cm.} \end{array} $$
Let the larger chord be $EF$ = 8 cm. Its perpendicular distance from the centre is $OM$. The radius $OF = OB$ = 5 cm.
$$ \begin{array}{l} \text{Half of the chord } MF = \frac{8}{2} = 4 \text{ cm.} \\ \text{In right-angled } \triangle OMF: \\ OM^2 + MF^2 = OF^2 \\ OM^2 + 4^2 = 5^2 \\ OM^2 + 16 = 25 \\ OM^2 = 25 - 16 = 9 \\ OM = \sqrt{9} = 3 \text{ cm.} \end{array} $$Answer: The distance of the other chord from the centre is 3 cm.
If the length of a chord of a circle is 48 cm and the distance of it from the centre is 7 cm. then let us write by calculating, the length of the chord which is 20 cm. distance away from the centre of the circle.
Explanation:
Let the first chord be $AB$ = 48 cm and its distance from the centre be $OD$ = 7 cm.
$$ \begin{array}{l} \text{Half of the chord } DB = \frac{48}{2} = 24 \text{ cm.} \\ \text{In right-angled } \triangle ODB\text{, radius } OB^2 = OD^2 + DB^2 \\ OB^2 = 7^2 + 24^2 = 49 + 576 = 625 \\ OB = \sqrt{625} = 25 \text{ cm.} \end{array} $$
Let the second chord be $EF$, and its distance from the centre is $OM$ = 20 cm. The radius $OF = OB$ = 25 cm.
$$ \begin{array}{l} \text{In right-angled } \triangle OMF: \\ MF^2 + OM^2 = OF^2 \\ MF^2 + 20^2 = 25^2 \\ MF^2 + 400 = 625 \\ MF^2 = 625 - 400 = 225 \\ MF = \sqrt{225} = 15 \text{ cm.} \end{array} $$The length of the chord $EF = 2 \times MF = 2 \times 15 = 30$ cm.
Answer: The length of the chord is 30 cm.
In the circle of adjoining figure with its centre at O, OP $\perp$ AB; if AB = 6 cm. and PC = 2 cm., then let us write by calculating, the length of radius of the circle.
Explanation:
From the given figure, $OP \perp AB$, $AB$ = 6 cm, and $PC$ = 2 cm.
Since the perpendicular from the centre bisects the chord, $PB = \frac{AB}{2} = \frac{6}{2} = 3$ cm.
Let the radius of the circle be $r$ cm. Therefore, $OB = OC = r$ cm.
The distance $OP = OC - PC = r - 2$ cm.
$$ \begin{array}{l} \text{In the right-angled } \triangle OPB\text{, by Pythagoras theorem:} \\ OP^2 + PB^2 = OB^2 \\ (r - 2)^2 + 3^2 = r^2 \\ r^2 - 4r + 4 + 9 = r^2 \\ -4r + 13 = 0 \\ 4r = 13 \\ r = \frac{13}{4} = 3.25 \text{ cm.} \end{array} $$Answer: The length of the radius of the circle is 3.25 cm.
A straight line intersects one of the two concentric circles at the points A and B and the other at the point C and D. I prove with reason that AC = DB.
Given : Let $O$ be the common centre of the two concentric circles. A straight line intersects the outer circle at points $A$ and $B$, and the inner circle at points $C$ and $D$.
To prove : $AC = DB$
Construction : From the centre $O$, the perpendicular $OM$ is drawn on the straight line $AB$.
Proof : For the outer circle, $OM$ is perpendicular on the chord $AB$.
$\therefore AM = MB$ [Since the perpendicular drawn from the centre to a chord, which is not a diameter, bisects the chord] ..... (i)
For the inner circle, $OM$ is perpendicular on the chord $CD$.
$\therefore CM = MD$ [By the same theorem] ..... (ii)
Subtracting equation (ii) from equation (i), we get :
$$ \begin{array}{l} AM - CM = MB - MD \\ \text{or, } AC = DB \end{array} $$$\therefore AC = DB$ [proved]
I prove that, the two intersecting chords of any circle can not bisect each other unless both of them are diameters of the circle.
Given : $AB$ and $CD$ are two intersecting chords of a circle with its centre at $O$, intersecting and bisecting each other at point $P$. i.e., $AP = PB$ and $CP = PD$.
To prove : $AB$ and $CD$ are diameters of the circle.
Construction : I join $O, P$.
Proof : Let us assume that the point $P$ is not the centre of the circle ($P \neq O$).
Since $P$ is the mid-point of chord $AB$, the line joining the centre $O$ to the mid-point $P$ is perpendicular to the chord $AB$.
$\therefore OP \perp AB$ i.e., $\angle OPB = 1$ right-angle.
Similarly, since $P$ is the mid-point of chord $CD$, the line $OP$ is perpendicular to $CD$.
$\therefore OP \perp CD$ i.e., $\angle OPD = 1$ right-angle.
Thus, through a single point $P$ on the line segment $OP$, two different perpendiculars $AB$ and $CD$ are drawn. This is geometrically impossible unless the lines $AB$ and $CD$ coincide. But $AB$ and $CD$ are two distinct intersecting chords.
$\therefore$ Our assumption is wrong. The point $P$ must coincide with the centre $O$.
Since the chords intersect and bisect each other at the centre $O$, both chords pass through the centre.
$\therefore$ Both $AB$ and $CD$ are diameters of the circle. [proved]
The two circles with centres X and Y intersect each other at the points A and B. A is joined with the mid-point 'S' of XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q. Let us prove that PA = AQ.
Given : Two circles with centres $X$ and $Y$ intersect at $A$ and $B$. $S$ is the mid-point of $XY$. A straight line passing through $A$ and perpendicular to $SA$ is drawn, which intersects the circles at $P$ and $Q$.
To prove : $PA = AQ$
Construction : From the centres $X$ and $Y$, the perpendiculars $XC$ and $YD$ are drawn on the line $PQ$.
Proof : $SA \perp PQ$ [Given]
$XC \perp PQ$ and $YD \perp PQ$ [By construction]
$\therefore XC \parallel SA \parallel YD$
In quadrilateral $XCDY$, the lines $XC, SA$, and $YD$ are parallel to each other. $S$ is the mid-point of $XY$ [given].
By the intercept theorem, if three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal.
$\therefore A$ is the mid-point of $CD$.
So, $CA = AD$ ..... (i)
In the circle with centre $X$, $XC \perp PA$.
$\therefore C$ is the mid-point of chord $PA$ [Since the perpendicular drawn from the centre to a chord bisects the chord]
So, $PA = 2CA$ ..... (ii)
In the circle with centre $Y$, $YD \perp AQ$.
$\therefore D$ is the mid-point of chord $AQ$.
So, $AQ = 2AD$ ..... (iii)
From (i), multiplying both sides by 2, we get $2CA = 2AD$.
$\therefore PA = AQ$ [Using (ii) and (iii)] [proved]
The two parallel chords AB and CD with the lengths of 10 cm and 24 cm in a circle are situated on the opposite sides of the centre. If the distance between two chords AB and CD is 17 cm., then let us write by calculating, the length of the radius of the circle.
Given : In a circle with its centre at $O$, $AB$ and $CD$ are two parallel chords situated on the opposite sides of the centre. $AB = 10$ cm and $CD = 24$ cm. The distance between the chords $AB$ and $CD$ is 17 cm.
To calculate : The length of the radius of the circle.
Construction : From the centre $O$, the perpendiculars $OE$ and $OF$ are drawn on the chords $AB$ and $CD$ respectively. $O, A$ and $O, C$ are joined.
Calculation : Let the length of the radius of the circle $OA = OC = r$ cm.
Let the distance of the chord $AB$ from the centre be $OE = x$ cm. Since $AB \parallel CD$, $E, O, F$ lie on a straight line, and the total distance is 17 cm.
$\therefore$ The distance of the chord $CD$ from the centre is $OF = (17 - x)$ cm.
$OE \perp AB \therefore AE = \frac{1}{2} AB = \frac{10}{2} = 5$ cm.
$OF \perp CD \therefore CF = \frac{1}{2} CD = \frac{24}{2} = 12$ cm.
In right-angled $\triangle AOE$, $OA^2 = OE^2 + AE^2$
$\therefore r^2 = x^2 + 5^2$ ..... (i)
In right-angled $\triangle COF$, $OC^2 = OF^2 + CF^2$
$\therefore r^2 = (17 - x)^2 + 12^2$ ..... (ii)
From (i) and (ii), we get :
$$ \begin{array}{l} x^2 + 25 = (17 - x)^2 + 144 \\ \text{or, } x^2 + 25 = 289 - 34x + x^2 + 144 \\ \text{or, } 34x = 289 + 144 - 25 \\ \text{or, } 34x = 408 \\ \text{or, } x = \frac{408}{34} = 12 \end{array} $$Putting the value of $x$ in (i), we get :
$$ \begin{array}{l} r^2 = (12)^2 + 5^2 \\ r^2 = 144 + 25 = 169 \\ \therefore r = \sqrt{169} = 13 \end{array} $$$\therefore$ The length of the radius of the circle is 13 cm.
The centres of two circles are P and Q; they intersect at the points A and B. The straight line parallel to the line-segment PQ through the point A intersects the two circles at the points C and D. I prove that, CD = 2PQ.
Given : Two circles with centres $P$ and $Q$ intersect each other at points $A$ and $B$. A straight line through $A$, parallel to $PQ$, intersects the two circles at $C$ and $D$. i.e., $PQ \parallel CD$.
To prove : $CD = 2PQ$
Construction : From the centres $P$ and $Q$, perpendiculars $PE$ and $QF$ are drawn on the line $CD$.
Proof : $PE \perp CD$ and $QF \perp CD$ [By construction]
$\therefore PE \parallel QF$.
Also, $PQ \parallel CD$ i.e., $PQ \parallel EF$ [Given]
In quadrilateral $PEFQ$, opposite sides are parallel and the angles are right-angles, so it is a rectangle.
$\therefore EF = PQ$ ..... (i)
In the circle with centre $P$, $PE \perp CA$.
$\therefore E$ is the mid-point of chord $CA$ [Since the perpendicular from centre bisects the chord].
So, $CA = 2EA$ ..... (ii)
In the circle with centre $Q$, $QF \perp AD$.
$\therefore F$ is the mid-point of chord $AD$.
So, $AD = 2AF$ ..... (iii)
Now, $CD = CA + AD$
$$ \begin{array}{l} \text{or, } CD = 2EA + 2AF \\ \text{or, } CD = 2(EA + AF) \\ \text{or, } CD = 2EF \end{array} $$From (i), substituting $EF$ with $PQ$, we get :
$\therefore CD = 2PQ$ [proved]
The two chords AB and AC of a circle are equal. I prove that, the bisector of $\angle BAC$ passes through the centre.
Given : In a circle with its centre at $O$, two chords $AB$ and $AC$ are equal i.e., $AB = AC$.
To prove : The bisector of $\angle BAC$ passes through the centre $O$.
Construction : $O, A$ ; $O, B$ and $O, C$ are joined.
Proof : In $\triangle AOB$ and $\triangle AOC$,
$AB = AC$ [Given]
$OA$ is common side.
$OB = OC$ [radii of same circle]
$\therefore \triangle AOB \cong \triangle AOC$ [By S-S-S axiom of congruency]
$\therefore \angle OAB = \angle OAC$ [Corresponding angles of congruent triangles]
This implies that the line segment $OA$ bisects the angle $\angle BAC$.
Since $OA$ is a line segment drawn from the vertex $A$ to the centre $O$, the angle bisector of $\angle BAC$ passes through the centre $O$.
$\therefore$ The bisector of $\angle BAC$ passes through the centre. [proved]
If the angle-bisector of two intersecting chords of a circle passes through its centre, then let me prove that the two chords are equal.
Given : In a circle with centre $O$, two chords $AB$ and $CD$ intersect each other at point $P$. The line segment $OP$ passes through the centre and is the angle-bisector of $\angle BPD$, meaning $\angle OPB = \angle OPD$.
To prove : $AB = CD$
Construction : Perpendiculars $OM$ and $ON$ are drawn from the centre $O$ to the chords $AB$ and $CD$ respectively.
Proof : In $\triangle OMP$ and $\triangle ONP$,
$\angle OMP = \angle ONP$ [Both are $90^\circ$ by construction]
$\angle OPM = \angle OPN$ [Since $OP$ is the angle bisector of $\angle BPD$, i.e., $\angle OPB = \angle OPD$]
$OP$ is the common side.
$\therefore \triangle OMP \cong \triangle ONP$ [By A-A-S axiom of congruency]
$\therefore OM = ON$ [Corresponding parts of congruent triangles]
We know that chords equidistant from the centre of a circle are equal in length.
Since the perpendicular distance $OM = ON$,
$\therefore AB = CD$ [proved]
I prove that, among two chords of a circle the length of the chord nearer to centre is greater than the length of the other.
Given : In a circle with its centre at $O$, $AB$ and $CD$ are two chords. $OM \perp AB$ and $ON \perp CD$. The chord $AB$ is nearer to the centre than the chord $CD$, which means $OM < ON$.
To prove : $AB > CD$
Construction : $O, B$ and $O, D$ are joined.
Proof : In right-angled $\triangle OMB$,
$$ \begin{array}{l} OM^2 + MB^2 = OB^2 \quad \text{[Pythagoras theorem]} \\ \therefore MB^2 = OB^2 - OM^2 \quad ..... \text{(i)} \end{array} $$In right-angled $\triangle OND$,
$$ \begin{array}{l} ON^2 + ND^2 = OD^2 \\ \therefore ND^2 = OD^2 - ON^2 \quad ..... \text{(ii)} \end{array} $$Now, $OB = OD$ [Radii of the same circle].
It is given that $OM < ON$, so $OM^2 < ON^2$.
Therefore, subtracting a smaller value ($OM^2$) from $OB^2$ will leave a greater remainder than subtracting a larger value ($ON^2$) from $OD^2$.
$$ \begin{array}{l} \therefore (OB^2 - OM^2) > (OD^2 - ON^2) \end{array} $$From (i) and (ii), we get :
$$ \begin{array}{l} MB^2 > ND^2 \\ \Rightarrow MB > ND \end{array} $$Since the perpendicular drawn from the centre to a chord bisects the chord, $AB = 2MB$ and $CD = 2ND$.
Multiplying both sides of the inequality by 2 :
$$ \begin{array}{l} 2MB > 2ND \\ \therefore AB > CD \quad \textbf{[proved]} \end{array} $$Let us write by proving which chord with the least length through any point in a circle.
Given : In a circle with centre $O$, let $M$ be any given point inside the circle. $PQ$ is a chord passing through $M$ such that $OM \perp PQ$. Let $RS$ be any other chord passing through $M$.
To prove : $PQ < RS$ (i.e., the chord perpendicular to the radius through the point is the shortest among all chords passing through that point).
Construction : A perpendicular $ON$ is drawn from the centre $O$ to the chord $RS$.
Proof : In $\triangle OMN$, $\angle ONM = 90^\circ$ [By construction].
Therefore, $\triangle OMN$ is a right-angled triangle where $OM$ is the hypotenuse.
Since the hypotenuse is the longest side of a right-angled triangle, we have :
$OM > ON$
We know that among two chords of a circle, the one nearer to the centre is greater in length, and the one farther from the centre is smaller in length.
Since the perpendicular distance $OM$ of chord $PQ$ is greater than the perpendicular distance $ON$ of chord $RS$,
$\therefore PQ < RS$
Thus, among all chords passing through a given point inside a circle, the chord which is perpendicular to the line joining the centre and the point has the least length. [proved]
📘 16. Very short answer type question
(A) M.C.Q.Explanation:
We know that equal chords of a circle subtend equal angles at the centre.
Given that chord \(AB =\) chord \(CD\), and \(\angle AOB = 60^\circ\).
Therefore, \(\angle COD = \angle AOB = 60^\circ\).
Correct Option: (c) \(60^\circ\)
Explanation:
Let the radius be \(r = 13\) cm, and the length of the chord be \(2l = 10\) cm. This implies \(l = 5\) cm.
Let \(d\) be the perpendicular distance from the centre to the chord. According to Pythagoras theorem:
\[\begin{array}{l} r^2 = d^2 + l^2 \\ 13^2 = d^2 + 5^2 \\ 169 = d^2 + 25 \\ d^2 = 169 - 25 = 144 \\ d = \sqrt{144} = 12 \text{ cm.} \end{array}\]Correct Option: (b) \(12\) cm.
Explanation:
We know that equal chords of a circle are equidistant from the centre.
Since chord \(AB =\) chord \(CD\), and the distance of \(AB\) from the centre is \(4\) cm, the distance of \(CD\) from the centre must also be \(4\) cm.
Correct Option: (b) \(4\) cm.
Explanation:
Since the two parallel chords \(AB\) and \(CD\) are equal in length (\(16\) cm), they must lie on opposite sides of the centre and be equidistant from it.
Let the perpendicular distance from the centre to each chord be \(d\).
Radius \(r = 10\) cm. Half of the chord length \(l = \frac{16}{2} = 8\) cm.
By Pythagoras theorem for one of the chords:
\[\begin{array}{l} r^2 = d^2 + l^2 \\ 10^2 = d^2 + 8^2 \\ 100 = d^2 + 64 \\ d^2 = 100 - 64 = 36 \\ d = \sqrt{36} = 6 \text{ cm.} \end{array}\]The total distance between the two parallel chords is \(2d = 2 \times 6 = 12\) cm.
Correct Option: (a) \(12\) cm.
Explanation:
Let's draw a perpendicular \(OE\) from the common centre \(O\) to the line intersecting the circles.
For the outer circle, \(CD\) is a chord. Since \(OE \perp CD\), \(E\) bisects \(CD\). Thus, \(CE = ED\).
For the inner circle, \(AB\) is a chord. Since \(OE \perp AB\), \(E\) bisects \(AB\). Thus, \(AE = EB\).
Now, we can express the lengths of \(AC\) and \(BD\) as:
\[\begin{array}{l} AC = CE - AE \\ BD = ED - EB \end{array}\]Since \(CE = ED\) and \(AE = EB\), it directly follows that:
\(AC = BD\)
Given \(AC = 5\) cm, therefore \(BD = 5\) cm.
Correct Option: (b) \(5\) cm.
📘 16. Very short answer type question
(B) True or FalseOnly one circle can be drawn through three collinear points.
Explanation:
Three collinear points lie on a single straight line. It is geometrically impossible to draw a circle that passes through all three of them. A circle can only be drawn through three non-collinear points.
Answer: False
The two circles $ABCDA$ and $ABCEA$ are same circle.
Explanation:
A unique circle is defined by any three non-collinear points. Since both circles pass through the same three points $A$, $B$, and $C$, they must be the exact same circle.
Answer: True
If two chords $AB$ and $AC$ of a circle with its centre $O$ are situated on the opposite sides of the radius $OA$, then $\angle OAB = \angle OAC$.
Explanation:
The angles $\angle OAB$ and $\angle OAC$ depend entirely on the lengths of the chords $AB$ and $AC$. If $AB \neq AC$, then the triangles $\triangle OAB$ and $\triangle OAC$ are not congruent, and their corresponding angles will not be equal. This statement is only true if the chords are equal in length, which is not stated.
Answer: False
📘 16. Very short answer type question
(C) Fill in the blanksExplanation:
The ratio of the chords is $1:1$, which means the chords are equal in length ($PQ = RS$). According to the theorem, equal chords of a circle subtend equal angles at the centre. Therefore, $\angle POQ = \angle ROS$, making their ratio $1:1$.
Answer: $1:1$
Explanation:
By the fundamental properties of a circle, the perpendicular bisector of any chord always passes through the centre of the circle.
Answer: passing through the centre
📘 17. Short answer type question (S.A.)
SolutionsTwo equal circles of radius 10 cm. intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centres of two circles.
Calculation:
Let \(M\) and \(N\) be the centres of the two equal circles intersecting at points \(A\) and \(B\).
Radius \(AM = AN = 10\) cm. Common chord \(AB = 12\) cm.
We know that the line joining the centres of two intersecting circles is the perpendicular bisector of their common chord. Let \(MN\) intersect \(AB\) at \(P\).
\[\begin{array}{l} \therefore AP = \frac{AB}{2} = \frac{12}{2} = 6 \text{ cm} \\ \text{In the right-angled } \triangle APM, \text{ by Pythagoras theorem:} \\ MP^2 + AP^2 = AM^2 \\ MP^2 + 6^2 = 10^2 \\ MP^2 = 100 - 36 = 64 \\ \therefore MP = \sqrt{64} = 8 \text{ cm} \end{array}\]Since the circles are equal, \(\triangle APN\) is identical to \(\triangle APM\), meaning \(PN = MP = 8\) cm.
The distance between the two centres is \(MN = MP + PN = 8 + 8 = 16\) cm.
Answer: The distance between the two centres is 16 cm.
AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of the circle is situated at the outside of the triangle ABC. If AB = AC = 6 cm., then let us calculate the length of the chord BC.
Calculation:
Let \(O\) be the centre of the circle. \(OA = OB = OC = 5\) cm (radii).
Given \(AB = AC = 6\) cm. Let the line \(OA\) intersect \(BC\) at \(M\). Since \(AB = AC\), \(OA\) is the perpendicular bisector of \(BC\).
Since the centre \(O\) lies outside the \(\triangle ABC\), the point \(M\) lies between \(A\) and \(O\).
Let \(OM = x\) cm. Then \(AM = OA - OM = (5 - x)\) cm.
\[\begin{array}{l} \text{In right-angled } \triangle ABM: \\ BM^2 = AB^2 - AM^2 \\ BM^2 = 6^2 - (5 - x)^2 \\ BM^2 = 36 - (25 - 10x + x^2) \\ BM^2 = 11 + 10x - x^2 \quad \dots (i) \\ \\ \text{In right-angled } \triangle OBM: \\ BM^2 = OB^2 - OM^2 \\ BM^2 = 5^2 - x^2 \\ BM^2 = 25 - x^2 \quad \dots (ii) \end{array}\]Equating (i) and (ii):
\[\begin{array}{l} 11 + 10x - x^2 = 25 - x^2 \\ 10x = 25 - 11 \\ 10x = 14 \\ x = 1.4 \text{ cm} \end{array}\]Now, substitute \(x = 1.4\) into equation (ii) to find \(BM\):
\[\begin{array}{l} BM^2 = 25 - (1.4)^2 \\ BM^2 = 25 - 1.96 \\ BM^2 = 23.04 \\ \therefore BM = \sqrt{23.04} = 4.8 \text{ cm} \end{array}\]The length of the chord \(BC = 2 \times BM = 2 \times 4.8 = 9.6\) cm.
Answer: The length of the chord \(BC\) is 9.6 cm.
The lengths of two chords AB and CD of a circle with its centre O are equal. If \(\angle AOB = 60^\circ\) and CD = 6 cm., then let us calculate the length of the radius of the circle.
Calculation:
In \(\triangle AOB\), \(OA = OB\) because they are both radii of the same circle.
Since angles opposite to equal sides are equal:
\[\begin{array}{l} \angle OAB = \angle OBA \\ \angle OAB + \angle OBA + \angle AOB = 180^\circ \\ 2\angle OAB + 60^\circ = 180^\circ \\ 2\angle OAB = 120^\circ \\ \angle OAB = \angle OBA = 60^\circ \end{array}\]Since all three angles of \(\triangle AOB\) are \(60^\circ\), it is an equilateral triangle.
Therefore, \(OA = OB = AB\).
We are given that chord \(AB =\) chord \(CD\), and \(CD = 6\) cm.
Thus, \(AB = 6\) cm.
Radius \(OA = AB = 6\) cm.
Answer: The length of the radius of the circle is 6 cm.
P is any point in a circle with its centre O. If the length of the radius is 5 cm. and OP = 3 cm., then let us determine the least length of the chord passing through the point P.
Calculation:
The chord with the least length passing through a specific point \(P\) inside a circle is the chord that is perpendicular to the radius passing through that point.
Let the chord be \(AB\). Then, \(OP \perp AB\).
Since the perpendicular from the centre to a chord bisects the chord, \(P\) is the mid-point of \(AB\). Thus, \(AB = 2 \times PB\).
\[\begin{array}{l} \text{In right-angled } \triangle OPB, \text{ by Pythagoras theorem:} \\ OP^2 + PB^2 = OB^2 \\ 3^2 + PB^2 = 5^2 \\ 9 + PB^2 = 25 \\ PB^2 = 25 - 9 = 16 \\ PB = \sqrt{16} = 4 \text{ cm} \end{array}\]The least length of the chord \(AB = 2 \times PB = 2 \times 4 = 8\) cm.
Answer: The least length of the chord passing through point \(P\) is 8 cm.
The two circles with their centres at P and Q intersect each other at the points A and B. Through the point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 5 cm., then let us determine the length of CD.
Calculation:
Let's draw perpendiculars \(PE\) and \(QF\) from the centres \(P\) and \(Q\) to the straight line \(CD\).
Since \(CD \parallel PQ\) (given) and \(PE \perp CD, QF \perp CD\), the figure \(PEFQ\) forms a rectangle.
Therefore, opposite sides are equal: \(EF = PQ = 5\) cm.
\[\begin{array}{l} \text{For the circle with centre } P, PE \perp CA. \\ \text{The perpendicular from the centre bisects the chord, so } CE = EA. \\ \therefore CA = 2EA \\ \\ \text{For the circle with centre } Q, QF \perp AD. \\ \text{Similarly, } AF = FD. \\ \therefore AD = 2AF \\ \\ \text{The total length of the chord } CD \text{ is } CA + AD. \\ CD = 2EA + 2AF \\ CD = 2(EA + AF) \\ \text{Since } EA + AF = EF, \\ CD = 2EF \end{array}\]Substitute \(EF = PQ = 5\) cm:
\(CD = 2 \times 5 = 10\) cm.
Answer: The length of \(CD\) is 10 cm.
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