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Practice each question, then open the answer or worked solution.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Consider the following diagram.
In ΔPOO’ and ΔQOO’
OP = OQ (Radius of circle 1)
O’P = O’Q (Radius of circle 2)
OO’ = OO’ (Common arm)
So, by SSS congruency, ΔPOO’ ≅ ΔQOO’
Thus, ∠OPO’ = ∠OQO’ (proved).
Two chords AB and CD of lengths 5 cm and 11 cm, respectively, of a circle, are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.
Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.
We know that AB bisects BM as the perpendicular from the centre bisects the chord.
Since AB = 5 so,
BM = AB/2 = 5/2
Similarly, ND = CD/2 = 11/2
Now, let ON be x.
So, OM = 6−x.
Consider ΔMOB,
OB2 = OM2+MB2
Or,
Consider ΔNOD,
OD2 = ON2 + ND2
Or
We know, OB = OD (radii)
From equation 1 and equation 2, we get
Now, from equation (2), we have,
OD2= 12 +(121/4)
Or OD = (5/2)×√5 cm
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance 4 cm from the centre, what is the distance of the other chord from the centre?
Consider the following diagram.
Here, AB and CD are 2 parallel chords. Now, join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
So, OM = 4 cm
MB = AB/2 = 3 cm
Consider ΔOMB.
OB2 = OM2+MB2
Or, OB = 5cm
Now, consider ΔOND.
OB = OD = 5 (Since they are the radii.)
ND = CD/2 = 4 cm
Now, OD2= ON2+ND2
Or, ON = 3 cm
Let the vertex of an angle ABC be located outside a circle, and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Consider the diagram.
Here AD = CE
We know any exterior angle of a triangle is equal to the sum of interior opposite angles.
So,
∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)
DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.
So,
∠DAE = (½)∠DOE ——————-(ii)
Similarly, ∠AEC = (½)∠AOC ——————-(iii)
Now, from equations (i), (ii), and (iii), we get
(½)∠DOE = ∠ABC+(½)∠AOC
Or, ∠ABC = (½)[∠DOE-∠AOC] (Hence, proved)
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
To prove: A circle drawn with Q as the centre will pass through A, B and O (i.e., QA = QB = QO).
Since all sides of a rhombus are equal,
AB = DC
Now, multiply (½) on both sides.
(½)AB = (½)DC
So, AQ = DP
BQ = DP
Since Q is the midpoint of AB,
AQ= BQ
Similarly,
RA = SB
Again, as PQ is drawn parallel to AD,
RA = QO
Now, as AQ = BQ and RA = QO, we get
QA = QB = QO (Hence, proved)
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.
So, ∠AEC+∠CBA = 180°
As ∠AEC and ∠AED are linear pairs,
∠AEC+∠AED = 180°
Or, ∠AED = ∠CBA … (1)
We know in a parallelogram, opposite angles are equal.
So, ∠ADE = ∠CBA … (2)
Now, from equations (1) and (2), we get
∠AED = ∠ADE
Now, AD and AE are angles opposite to equal sides of a triangle.
∴ AD = AE (proved)
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Here, chords AB and CD intersect each other at O.
Consider ΔAOB and ΔCOD.
∠AOB = ∠COD (They are vertically opposite angles.)
OB = OD (Given in the question.)
OA = OC (Given in the question.)
So, by SAS congruency, ΔAOB ≅ ΔCOD
Also, AB = CD (By CPCT)
Similarly, ΔAOD ≅ ΔCOB
Or, AD = CB (By CPCT)
In quadrilateral ACBD, opposite sides are equal.
So, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
So, ∠A = ∠C
Also, as ABCD is a cyclic quadrilateral,
∠A+∠C = 180°
⇒∠A+∠A = 180°
Or, ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, so, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the triangle DEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.
Consider the following diagram.
Here, ABC is inscribed in a circle with centre O, and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F, respectively.
Now, join DE, EF and FD.
As angles in the same segment are equal, so,
∠EDA = ∠FCA ————-(i)
∠FDA = ∠EBA ————-(i)
By adding equations (i) and (ii), we get
∠FDA+∠EDA = ∠FCA+∠EBA
Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B
We know, ∠A +∠B+∠C = 180°
So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]
∠FDE = [90-(∠A/2)]
In a similar way,
∠FED = [90° -(∠B/2)] °
And,
∠EFD = [90° -(∠C/2)] °
Two congruent circles intersect each other at points A and B. Through A, any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
The diagram will be
Here, ∠APB = ∠AQB (as AB is the common chord in both the congruent circles.)
Now, consider ΔBPQ.
∠APB = ∠AQB
So, the angles are opposite to equal sides of a triangle.
∴ BQ = BP
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Consider this diagram.
Here, join BE and CE.
Now, since AE is the bisector of ∠BAC,
∠BAE = ∠CAE
Also,
∴arc BE = arc EC
This implies chord BE = chord EC
Now, consider triangles ΔBDE and ΔCDE.
DE = DE (It is the common side)
BD = CD (It is given in the question)
BE = CE (Already proved)
So, by SSS congruency, ΔBDE ≅ ΔCDE.
Thus, ∴∠BDE = ∠CDE
We know, ∠BDE = ∠CDE = 180°
Or, ∠BDE = ∠CDE = 90°
∴ DE ⊥ BC (Hence, proved).