Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Given: Two circles with Centre A and B circle intersects at C and D.
To prove: ∠ACB=∠ADB
Construction: Join AD,BC and BD
Proof: In ∆ACB and ∆ADB
AC=AD ——–[radii of the same circle]
BC=BD ———[radii of the same circle]
∴∆ACB≅∆ADB ——— [by S.S.S]
Question 2. 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 cm, find the radius of the circle.
Let O be the centre of circle and be r cm
Given: AB=5cm, CD =11cm
Construction: Draw OM perpendicular AB and OL perpendicular CD.
Because OM perpendicular AB and OL perpendicular CD and AB||CD.
∴Points O,L, and M are collinear, than ∠M=6cm
Join AO and CO
OL=1/2CD=1/2*11=5.5cm —–[perpendicular from bisects the chord]
AM=1/2AB=1/2*5=2.3cm —–[perpendicular from bisects the chord]
Now, In right ∆DLC
Now in right ∆OMA
Now equating equation 1 and 2
Putting value of x in equation 1
Radius of circle is 5.6cm.
Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the centre?
Let AB and CD are|| chord of circle with centre O which AB=6cm and CD=8cm and radius of circle =r cm.
Construction: Draw OP perpendicular AB and OM perpendicular CD.
Because AB||CD and OP perpendicular AB and OM perpendicular CD therefore. Point O, M and P are collinear.
Clearly, OP=4cm ———-[According to question]
P is midpoint of AB.
M is midpoint of AB.
Join AO and CO
Now in Right ∆OPA,
Now in ∆OMC
∴Therefore distance of the other chord from the centre is 3cm
Question 4. 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.
Give: Vertex B of ∆ABC lie outside the circle,chord AD=CE
To prove: ∠ABC=1/2(∠DOE-∠AOC)
Construction: Join AE
Solution: Chord DE subtends ∠DOE at the center and ∠DAE at point A on the circle.
chords AC subtends ∠AOC at the centre and ∠AEC at point
In ∆ ABE,∠DAE is exterior angle
Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Given: A rhombus ABCD in which O is intersecting point of diagonals AC and BD.
A circle is drawn taking CD as diameter.
To prove: circle points through O or Lies on the circles.
Proof: In rhombus ABCD,
∠DOA=90° ——–[diagonals of rhombus intersect at 90°] 1
∠COD=90° ——–[angle made in segment O is right angle] 2
From 1 and 2
O lies on the circle.
Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
ABCD is a ||gm. The circle through A,B and C intersect at E.
To prove: AE=AD
Proof: Here ABCE is a cyclic quadrilateral
∠2+∠4=180° —–[sum of opposite is of a cyclic quadrilateral is 180°]
Now ∠4+∠6=180°-∠6 ———2
From 1 and 2
Also ∠2=∠5 ———[opposite angles of ||gm are equal] —–4
From 3 and 4
Now, In ∆ADE,
∴AE=AD ——[sides apposite to equal angles in a∆ are equal]
Question 7. 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.
Given: Two chords AC and BD bisects each other i.e OA=OC,OB=OD
To prove: In ∆AOB and COB
∠AOB=∠COD ——[vertically opposite angle]
AB=CD ——[C.P.C.T.] 1
similarly ∆AOD≅∆COB (S.A.S)
AD=CD (C.P.C.T.) 2
From 1 and 2 ABCD is a ||gm
Since, ABCD is cyclic quadrilateral
∴ ∠A and ∠B lies in a semicircle
→ AC and BD are diameter of circle.
ii) Since ABCD is a ||gm and ∠A=90°
∴ ABCD is a rectangle.
Question 8. 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° – 1 2 A, 90° – 1 2 B and 90° – 1 2 C.
Given:∆ABC and it circum-circle AD,BE and CF are bisectors of ∠A,∠B and ∠C
To proof:∠ D=90°-1/2∠A , ∠E=90°-1/2∠B , ∠F=90°-1/2∠C
Construction: Join AE and AF.
Solution: ∠ADE=∠ABE ———-1 [angle in the same segment are equal]
∠ADF=∠ACF ———–2 [angle in the same segment are equal]
Adding 1 and 2
∠D=1/2∠B+1/2∠C ——[BC and CF are bisector of ∠B & ∠c]
Question 9. 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.
Given: two congruent circles which intersect at A and B.
PAB is a line segment
To prove: BA=BQ
Construction: join AB
Proof: AB is a common chord of both the congruent circle.
Segment of both circles will be equal
Now, in ∆ BPQ,
BP=BQ ——[sides opposite to equal angles are equal]
Question 10. 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.
Given: A ∆ABC, in which AD is angle bisector of ∠A and OD is ⊥ bisector of BC.
To prove: D lies on circumcircle.
Construction: Join OB and OC
Proof: Since BC subtends ∠BAC at A on the remaining of the circle.
Now, In ∆BOE and ∆ COE
BO=OE ——–(radii of the same circle)
2∠1=2∠BAC ———- (from 1)
∠BOD=2∠BAD [AD is bisector of ∠BAC]
This is possible only if BD is chord of the circle.
D lies on the circle.