435 + Interview Questions in CIRCLE IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Let O(0, 0) and (0, 1) be two fixed points, then the locus of a point P such that the perimeter of `DeltaAOP` is 4, isA. `8x^(2)-9y^(2)+9y=18`B. `9x^(2)-8y^(2)+8y=16`C. `9x^(2)+8y^(2)-8y=16`D. `8x^(2)+9y^(2)-9y=18`
Answer» Correct Answer - C Given vertices of `DeltaAOP` are O(0, 0) and A(0, 1) Let the coordinates of point P are (x, y) Clearly, perimeter = OA + AP + OP = 4 (given) `rArrsqrt((0-0)^(2)+(0- 1)^(2))+sqrt((0- x)^(2) + (1-y)^(2))+sqrt(x^(2) +y^(2))=4` `rArr 1 + sqrt(x^(2)+(y-1)^(2))+sqrt(x^(2)+y^(2))=4` `rArr sqrt(x^(2) + y^(2) - 2y +1) +sqrt(x^(2) + y^(2))=3` `rArrx^(2)+y^(2)-2y + 1=3-sqrt(x^(2)+y^(2))` `rArrx^(2)+y^(2)-2y +1=9+x^(2)+y^(2)-6sqrt(x^(2)+y^(2))` [squaring both sides ] `rArr 1- 2y = 9 - 6sqrt(x^(2)+y^(2))` `rArr6sqrt(x^(2) + y^(2)) = 2y + 8 ` `rArr3sqrt(x^(2) + y^(2)) = y + 4 ` `rArr 9(x^(2) + y^(2)) = (y+4)^(2)` [ squaring both sides] `rArr 9x^(2)+9y^(2)=y^(2)+8y + 16` `rArr 9x^(2)+8y^(2)-8y=16` Thus, the locus of point P(x,y) is `9x^(2)+8y^(2)-8y = 16`

Discussion

152.	Find the equation of a circle with: Centre at (2, -3) and radius 5
Answer» The equation of a circle with centre at (h, k) and radius ‘r’ is given by (x – h)² + (y – k)² = r ² Here, h = 2, k = -3 and r = 5 The required equation of the circle is (x – 2)² + [y – (-3)]² = 5² ⇒ (x – 2)² + (y + 3)² = 25 ⇒ x² – 4x + 4 + y² + 6y + 9 – 25 = 0 ⇒ x² + y² – 4x + 6y – 12 = 0

152.

Find the equation of a circle with: Centre at (2, -3) and radius 5

Answer»

The equation of a circle with centre at (h, k) and radius ‘r’ is given by

(x – h)² + (y – k)² = r ²

Here, h = 2, k = -3 and r = 5

The required equation of the circle is

(x – 2)² + [y – (-3)]² = 5²

⇒ (x – 2)² + (y + 3)² = 25

⇒ x² – 4x + 4 + y² + 6y + 9 – 25 = 0

⇒ x² + y² – 4x + 6y – 12 = 0

Discussion

153.	Find the equation of a circle with: centre at (-3, -2) and radius 6.
Answer» The equation of a circle with centre at (h, k) and radius ‘r’ is given by (x – h)² + (y – k)² = r² Here, h = -3, k = -2 and r = 6 ∴ The required equation of the circle is [x – (-3)]² + [y – (-2)]² = 6² ⇒ (x + 3)² + (y + 2)² = 36 ⇒ x² + 6x + 9 + y² + 4y + 4 – 36 = 0 ⇒ x² + y² + 6x + 4y – 23 = 0

153.

Find the equation of a circle with: centre at (-3, -2) and radius 6.

Answer»

The equation of a circle with centre at (h, k) and radius ‘r’ is given by

(x – h)² + (y – k)² = r²

Here, h = -3, k = -2 and r = 6

∴ The required equation of the circle is

[x – (-3)]² + [y – (-2)]² = 6²

⇒ (x + 3)² + (y + 2)² = 36

⇒ x² + 6x + 9 + y² + 4y + 4 – 36 = 0

⇒ x² + y² + 6x + 4y – 23 = 0

Discussion

154.	Find the equation of a circle with : centre at origin and radius 4.
Answer» The equation of a circle with centre at origin and radius ‘r’ is given by x² + y² = r² Here, r = 4 ∴ The required equation of the circle is x² + y² = 4² i.e., x² + y² = 16.

154.

Find the equation of a circle with : centre at origin and radius 4.

Answer»

The equation of a circle with centre at origin and radius ‘r’ is given by

x² + y² = r²

Here, r = 4

∴ The required equation of the circle is

x² + y² = 4² i.e., x² + y² = 16.

Discussion

155.	P is a variable point on the line `L=0` . Tangents are drawn to the circles `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogram PQSR is completed. If `P-= (6,8)`, then the area of `Delta QRS` isA. `(3sqrt(6))/(25)` sq. unitsB. `(3sqrt(24))/(25)` sq. unitsC. `(48sqrt(6))/(25)` sq. unitsD. `(192sqrt(6))/(25)` sq. units
Answer» Correct Answer - 4 `P -= (6,8)` Thereforem the equation of QR (chord of contact ) is `6x+8y=4` or `3x+4y-2=0 ` `:. PM =(48)/(50` and `PQ =sqrt(96)` `QM =sqrt(96-(48^(2))/(25))=sqrt((96)/(25))` `:. QR = 2 sqrt((96)/(25))` `:. ` Area of `Delta PQR =(1)/(2) xx PM xx QR =(192 sqrt(6))/(25)` PQRS is a rhombus. Therefore, Area of `Delta RS =` Area of `Delta PQR = (192sqrt(6))/(25)`

Discussion

156.	P is a variable point on the line `L=0` . Tangents are drawn to the circles `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogram PQSR is completed. If `P -=(3,4)`, then the coordinates of S areA. `(-46//25,63//25)`B. `(-51//25,-68//25)`C. `(-46//25,68//25)`D. `(-68//25,51//25)`
Answer» Correct Answer - 2 As `P -= (3,4)` , the equation of QR is `3x+4y=4 ` (1) Let ` S -= (x_(1),y_(1))`. S is the mirror image of P w.r.t. (1). Then, `(x_(1)-3)/(3)=(y_(1)-4)/(4)= (-2(3xx3+4xx4-4))/(3^(2)+4^(2))=-(42)/(25)` `:. x_(1)=-(51)/(25),y_(1)=-(68)/(25)` or `S-=(-(51)/(25),-(68)/(25))`

Discussion

157.	P is a variable point on the line L=0. Tangents are drawn to the circle `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogran PQSR is completed. If P-=(6,8) then area of `DeltaQRS` is `(192)/(25)sqrtlamda` sq units. The value of `lamda` isA. 2B. 3C. 5D. 6
Answer» Correct Answer - D

Discussion

158.	P is a variable point on the line L=0. Tangents are drawn to the circle `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogran PQSR is completed. If p-=(3,4), then the coordinates of S areA. `(-(46)/(25),(63)/(25))`B. `(-(51)/(25),-(68)/(25))`C. `(-(46)/(25),(68)/(25))`D. `(-(68)/(25),(51)/(25))`
Answer» Correct Answer - B

Discussion

159.	Equation of the circumcircle of a triangle formed by the lines `L_(1)=0,L_(2)=0andL_(3)=0` can be written as `L_(1)L_(2)+lamdaL_(2)L_(3)+muL_(3)L_(1)=0`, where `lamdaandmu` are such that coefficient of `x^(2)` =coefficient of `y^(2)` and coefficient of xy=0. `L_(1)L_(2)^(2)+lamdaL_(2)L_(3)^(2)+muL_(1)^(2)=0` representsA. a curve passing through point of interesection of `L_(1)=0,L_(2)=0 andL_(3)=0`B. a circle is coefficient of `x^(2)=` coefficient of `y^(2)` and coefficient of xy=0C. a parabolaD. pair of straight lines
Answer» Correct Answer - A

Discussion

160.	Equation of the circumcircle of a triangle formed by the lines `L_(1)=0,L_(2)=0andL_(3)=0` can be written as `L_(1)L_(2)+lamdaL_(2)L_(3)+muL_(3)L_(1)=0`, where `lamdaandmu` are such that coefficient of `x^(2)` =coefficient of `y^(2)` and coefficient of xy=0. `L_(1)L_(2)^(2)+lamdaL_(2)L_(3)^(2)+muL_(1)^(2)=0` representsA. `lamdaL_(1)L_(4)+muL_(2)L_(3)=0`B. `lamdaL_(1)L_(3)+muL_(2)L_(4)=0`C. `lamdaL_(1)L_(2)+muL_(3)L_(4)=0`D. `lamdaL_(1)^(2)L_(3)+muL_(2)^(2)L_(4)=0`
Answer» Correct Answer - C

Discussion

161.	Equation of the circumcircle of a triangle formed by the lines `L_(1)=0,L_(2)=0andL_(3)=0` can be written as `L_(1)L_(2)+lamdaL_(2)L_(3)+muL_(3)L_(1)=0`, where `lamdaandmu` are such that coefficient of `x^(2)` =coefficient of `y^(2)` and coefficient of xy=0. If `L_(1)L_(2)+lamdaL_(2)L_(3)+muL_(3)L_(1)=0` is such that `mu=0andlamda` is non-zero, then it representsA. a parabolaB. a pair of straight linesC. a circleD. an ellipse
Answer» Correct Answer - B

Discussion

162.	In the adjoining figure, AD is the diameter of the circle and `angleBCD=140^@`. Find the value of `angleADB`.
Answer» Correct Answer - `50^@`

Discussion

163.	In the given figure, find `angle ABC`.
Answer» Correct Answer - `70^@`

Discussion

164.	In the given figure , find the length of chord AB .
Answer» Correct Answer - 24 cm

Discussion

165.	Find `angle AOB` in the given figure.
Answer» Correct Answer - `100^@`

Discussion

166.	In the given figure, find `angle AOB`.
Answer» Correct Answer - `160^@`

Discussion

167.	Find `angle A` in the given figure.
Answer» Correct Answer - `100^@`

Discussion

168.	In the given figure, find `angleD` and `angleB`.
Answer» Correct Answer - `72^@,108^@`

Discussion

169.	In the given figure , find `angleDCP`.
Answer» Correct Answer - `70^@`

Discussion

170.	In the adjoining figure, AOB is the diameter of the circle. If `angleABP=45^@`, then find the value of `angle PQB`.
Answer» Correct Answer - `45^@`

Discussion

171.	In the adjoining figure, `O` is the centre of the circle. If `anglePAO=15^@` and `anglePBO=30^@`, then find the value of `angleAOB`.
Answer» Correct Answer - `90^@`

Discussion

172.	find the equation of circle whose centre `(2, 4 ) ` and radius 5
Answer» Correct Answer - ` x ^(2) + y ^(2) - 4x - 8y - 5= 0`

Discussion

173.	Find the centre and radius of each of the following circles : (i) ` (x - 3)^(2) + (y- 1) ^(2) = 9` (ii) ` (x - (1)/(2) ) ^(2) + ( y + (1)/(3) ) ^(2) = (1)/(16)` (iii) ` (x + 5) ^(2) + ( y- 3 ) ^(2) = 20` (iv)` x ^(2) + (y- 1 ) ^(2) = 2 `
Answer» Correct Answer - (i) Centre ` (3, 1 )`, radius = 3 (ii) Centre ` ((1)/(2), (-1)/(3))`, radius `= (1)/(4)` (iii) Centre ` (-5, 3 )` , radius = ` 2sqr5` (iv) Centre `(0, 1 )`, radius `= sqrt 2`

Discussion

174.	Show that the points `(5, 5 ), (6, 4 ), (-2, 4) and (7, 1 )` are concyclic, i.e., all lie on the same circle. Find the equation, centre and radius of this circle.
Answer» Let the equation of the circle passing through the points ` (5, 5 ), (6, 4 ) and (7, 1 )` be given by ` x ^(2) + y ^(2) + 2gx + 2f y + c = 0" " ` …(i) Then, each of these points must satisfy (i) ` 25 + 25 + 10 g + 10 f + c = 0 rArr 10 g + 10 f + c = - 50" " `... (ii) ` 36 + 16 + 12 g + 8 f + c = 0 rArr 12 g + 8f + c = - 52 " " `... (iii) ` 49 + 1 + 14 g + 2 f + c =0 rArr 14 g + 2 f + c = - 50 " " `... (iv) Substracting (ii) from (iv) , we get ` 4 g - 8f = 0 rArr g - 2 f = 0 " " `... (v) Substracting (iii) from (iv), we get ` 2 g - 6f = 2 rArr g -3 f = 1 " " ` ... (vi) Solving (v) and (vi), we get ` g = - 2 and f = -1 ` Putting these values in (ii), we get ` c = -20` Putting ` g = - 2 , f = - 1 and c = - 20 ` in (i), we get ` x ^(2) + y ^(2) - 4x - 2y - 20 = 0" " ` ... (vii) This is the equation of the circle passing through the points `(5, 5 ), (6, 4 ) and (7, 1)`. Putting ` x = - 2 and y = 4 ` in (vii), we get LHS `= 4 + 16 + 8 -8 - 20 = 0 = RHS`. This shows that the point ` (-2, 4)` also lies on (vii) Hence, the points `(5, 5 ), (6, 4 ), (-2, 4) and (7, 1 )` all lies on the same circle, given by (vii) Centre of this centre ` = (-g, - f ) = (2, 1 )` Radius of this circle = ` sqrt (g ^(2) + f ^(2) - c ) = sqrt ((-2 ) ^(2) + (-1) ^(2) + 20 )` ` " " = sqrt ( 25) = 5 ` units.

Discussion

175.	Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____. (A) on the centre (B) inside the circle (C) outside the circle (D) on the circle
Answer» (C) outside the circle l(OP) > radius ∴ Point P lies in the exterior of the circle.

175.

Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____. (A) on the centre (B) inside the circle (C) outside the circle (D) on the circle

Answer»

(C) outside the circle

l(OP) > radius

∴ Point P lies in the exterior of the circle.

Discussion

176.	find the equation of circle whose centre ` (-3, - 2)` and radius 6
Answer» Correct Answer - ` x ^(2) + y ^(2) + 6x + 4y - 23 = 0`

Discussion

177.	Show that the points `A(1, 0), B(2, -7), C(8, 1)` and `D(9,-6)` all lie on the same circle. Find the equation of this circle, its centre and radius.
Answer» Correct Answer - ` x ^(2) + y^(2) - 10x + 6y + 9 = 0 `, centre `C(5, - 3)` and radius = 5

Discussion

178.	centre at the origin and radius 4
Answer» Correct Answer - ` x ^(2) + y ^(2) - 16 = 0 `

Discussion

179.	centre ` (a, a)` and radius ` sqrt2`
Answer» Correct Answer - ` x ^(2) + y ^(2) - 2ax - 2ay = 0`

Discussion

180.	Find the equation of the circle with centre : `(-a , b)`and radius `sqrt(a^2-b^2)`.
Answer» Correct Answer - ` x ^(2) + y ^(2) + 2ax + 2 by + 2b ^(2) = 0 `

Discussion

181.	If the equation `p x^2+(2-q)x y+3y^2-6q x+30 y+6q=0`represents a circle, then find the values of `pa n dqdot`
Answer» In the equation of circle, the xy term does not occur and the coefficients of `x^(2)` and `y^(2)` are equal. Therefore, `2-q=0` or `q=2` and `p=3` Also, for these values of p and q, the circle is real.

Discussion

182.	Find the equation of the circle with: Centre `(a cosalpha, a s inalpha)`and radius `adot`
Answer» Correct Answer - ` x ^(2) + y ^(2) - 2ax cos alpha - 2ay sin alpha = 0 `

Discussion

183.	A circle is inscribed in an equilateral triangle of side `adot`The area of any square inscribed in this circle is ______.
Answer» Correct Answer - `(a^(2))/(6)`sq unit In an equilateral triangle, the radius of incircle `=(1)/(3)xx` median of the triangle `=(1)/(3)sqrt(a^(2)-(a^(2))/(4))=(1)/(3)sqrt((4a^(2)-a^(2))/(4))=(a)/(2sqrt3)` therefore ,area of the square inscribed in this circle `=2("radius of circle")^(2) = (2a^(2))/(4*3)= (a^(2))/(6)` sq unit

Discussion

184.	If `x^2+y^2-2x+2a y+a+3=0`represents the real circle with nonzero radius, then find the values of`adot`A. `ain(-oo,-1)`B. `ain(-1,2)`C. `ain(2,oo)`D. `ain(-oo,-1)uu(2,oo)`
Answer» Correct Answer - D

Discussion

185.	If `x^2+y^2-2x+2a y+a+3=0`represents the real circle with nonzero radius, then find the values of`adot`
Answer» For circle `x^(2)+y^(2)-2x+2a+a+3=0` , we have `g= -1,f=a` and `c=a+3` For real circle with nonzero radius, or `1+a^(2)-(a+3)gt0` or `a^(2)-a-2gt0` or `(a-2)(a+1)gt0` i.e., `alt -1` or `a gt 2`

Discussion

186.	In a cyclic `squareABCD` , twice the measure of `angleA` is thrice the measure of `angleC` . Find the measure of `angleC` .A. 36B. 72C. 90D. 108
Answer» Correct Answer - B

Discussion

187.	A point `P`moves in such a way that the ratio of its distance from two coplanarpoints is always a fixed number `(!=1)`. Then, identify the locus of the point.
Answer» Let two coplanar points be A(0,0) and B9a,0). Let P(x,y) be the variable point. According to question, `(AP)/(BP)=lambda` `:. (sqrt(x^(2)+y^(2)))/(sqrt((x-a)^(2)+y^(2)))=lambda`, (where `lambda` is any fixed number and `lambda cancel(=)1)` or `x^(2)+y^(2)+((lambda^(2))/(lambda^(2)-1))(a^(2)-2ax)=0` which is the equation of a circle.

Discussion

188.	In the adjoining figure, `O` is the centre of the circle.If `angleBAC=40^@`, then find the value of `angleADC`.
Answer» Correct Answer - `130^@`

Discussion

189.	ABCD is a cyclic trapezium in which, `AD\| \|BC` and `angleB=70^@`. Find its remaining angles,
Answer» Correct Answer - `angleA=110^@`,angleC=70^@`,angleD=110^@`

Discussion

190.	In the adjoining figure , `O` is the center of the circle. If `angleBAD=30^(@)` , then find the values of x,y and z.
Answer» Here, `angle BED =angle BAD ` (angles of a seagment) `rArr z=30^@` Arc BCD subtends angles at centre`= angle BOD =Y` remaining circle ` =angle BAD =30^@` `therefore angle BOD =2xxangle BAD` `rArr y=2xx30^@=60^@` ABCD is a cyclic quadrilateral. `therefore angle BCD +angle BAD =180^@` `rArrx+30^@=180^@` `rArrx=180^@-30^@` `rArrx=150^@` `therefore x=150^@, y=60^@, [email protected]`

Discussion

191.	The angle made by the tangent and the radius made at the point of contact isA. `0^(@)`B. `45^(@)`C. `90^(@)`D. `75^(@)`
Answer» Correct Answer - C

Discussion

192.	Find the image of the circle `x^2+y^2-2x+4y-4=0`in the line `2x-3y+5=0`
Answer» The image of the circle in the line will be circle having same radius, or given circle, centre is C(1,-2) and radius is `sqrt((-1)^(2)+(2)^(2)-(-4))=3` Let image of C in the line be `C_(1)(h,k) `. `:. (h-1)/(2)=(k+2)/(-3)=((-2(2(1))-3(2)+5))/((2)^(2)+(-3)^(2))= -2` `:. h= -3 ` and `k=4` Thus, centre of the image is `C_(1)(-3,4)` Hence, equation of the image circle is `(x+3)^(2)+(y-4)^(2)=9`

Discussion

193.	Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that `/_A C P =/_Q C D`.
Answer» Given : Two circles intersect at two points B and C. Through B two line segments ABD and PBQ are drawn to intersect the circles at A,D and P,Q, respectively. To prove: `angle ACP:angle QCD` Proof: In circle I, `angle ACP=angleABP` (angles in the same segment)....1) In circle II, ` angle QCD=angle QBD` (angles in the same segment )......(2) `angle ABP=angle QBD` (vertically opposite angles ) From eqs. (1) and (2), we get `angle ACP=angle QCD`

Discussion

194.	From the information given in the figure, find the measure of `/_AEC `. A. `42^(@)`B. `30^(@)`C. `36^(@)`D. `72^(@)`
Answer» Correct Answer - C

Discussion

195.	In the following, write true/false and if possible give reason for your answer.(i) AB = 3 cm and CD = 4 cm are two chords of a circle. Angles subtended by these chords at center are respectively 70° and 50°.(ii) Two chords of length 10 cm and 8 cm are at a distance from center 8 cm and 5 cm respectively.(iii) If two chords AB and CD of a circle are at same distance 4 cm from center then AB = CD.(iv) Two congruent circles having centers O and O’ intersect each other at two points A and B, then ∠AOB = ∠AO’B.(v) A circle can be drawn from three collinear points.(vi) A circle of 4 cm radius, passing through two points A and B can be drawn of AB = 8 cm.
Answer» (i) False, because longer chord subtends larger angle at the center, where as shorter chord subtends smaller angle at the center. (ii) False, because larger chords are closer from the center. (iii) True, because equal chords are equidistant from the center. (iv) True, because chords of congruent circles, subtend equal angle at the center. (v) False, because a circle cannot be drawn though three collinear points. (vi) True, because AB is a diameter.

195.

In the following, write true/false and if possible give reason for your answer.(i) AB = 3 cm and CD = 4 cm are two chords of a circle. Angles subtended by these chords at center are respectively 70° and 50°.(ii) Two chords of length 10 cm and 8 cm are at a distance from center 8 cm and 5 cm respectively.(iii) If two chords AB and CD of a circle are at same distance 4 cm from center then AB = CD.(iv) Two congruent circles having centers O and O’ intersect each other at two points A and B, then ∠AOB = ∠AO’B.(v) A circle can be drawn from three collinear points.(vi) A circle of 4 cm radius, passing through two points A and B can be drawn of AB = 8 cm.

Answer»

(i) False, because longer chord subtends larger angle at the center, where as shorter chord subtends smaller angle at the center.

(ii) False, because larger chords are closer from the center.

(iii) True, because equal chords are equidistant from the center.

(iv) True, because chords of congruent circles, subtend equal angle at the center.

(v) False, because a circle cannot be drawn though three collinear points.

(vi) True, because AB is a diameter.

Discussion

196.	A chord of 24 cm length is drawn in a circle of radius 13 cm. The distance of chord from center of circle is –(a) 12 cm(b) 5 cm(c) 6.5 cm(d) 12 cm
Answer» Answer is (b) 5 cm AM = BM = \(\frac { 1 }{ 2 }\) × AM = \(\frac { 1 }{ 2 }\) × 24 = 12 cm In right ∆OAM OA² = AM² + OM² (13)² = (12)² + (OM)² OM² = (13)² – (12)² = 169 – 144 OM² = 25 OM = \(\sqrt { 25 }\) = 5 cm Thus, distance of chord from center of circle = 5 cm.

196.

A chord of 24 cm length is drawn in a circle of radius 13 cm. The distance of chord from center of circle is –(a) 12 cm(b) 5 cm(c) 6.5 cm(d) 12 cm

Answer»

Answer is (b) 5 cm

AM = BM = \(\frac { 1 }{ 2 }\) × AM = \(\frac { 1 }{ 2 }\) × 24 = 12 cm

In right ∆OAM

OA² = AM² + OM²

(13)² = (12)² + (OM)²

OM² = (13)² – (12)² = 169 – 144

OM² = 25

OM = \(\sqrt { 25 }\) = 5 cm

Thus, distance of chord from center of circle = 5 cm.

Discussion

197.	In given fig., O is center of the circle. Chord AB = CD and ∠OBA = 40°, then ∠COD will be :(A) 100°(B) 80°(C) 180°(D) 90°
Answer» Answer is (A) 100° Given : AB = CD Radius of circle = OB = OD ∴ ∠OBA = ∠ODC = 40° ∆OCD in OD = OC (Radius of circle) ∴ ∆OCD is an isosceles triangle. ∠OCD = ∠ODC = 40° In ∆OCD, ∠ODC + ∠OCD + ∠COD = 180° 40° + 40° + ∠CQD = 180° ∠COD = 180° – 40° – 40° ∠COD = 180° – 80° ∠COD = 100°

197.

In given fig., O is center of the circle. Chord AB = CD and ∠OBA = 40°, then ∠COD will be :(A) 100°(B) 80°(C) 180°(D) 90°

Answer»

Answer is (A) 100°

Given : AB = CD

Radius of circle = OB = OD

∴ ∠OBA = ∠ODC = 40°

∆OCD in OD = OC (Radius of circle)

∴ ∆OCD is an isosceles triangle.

∠OCD = ∠ODC = 40°

In ∆OCD,

∠ODC + ∠OCD + ∠COD = 180°

40° + 40° + ∠CQD = 180°

∠COD = 180° – 40° – 40°

∠COD = 180° – 80°

∠COD = 100°

Discussion

198.	In the adjoining figure, ABCD is a cyclic quadrilateral.If side BC is produced upto point E and `angleDAB=95^@`, then find the value of `angleDCE`.
Answer» Correct Answer - `95^@`

Discussion

199.	In the adjoining figure, ABCD is a cyclic quadrilateral and `AB` is the diameter of the circle. If `angleAPC=120^@`, then find the value of `angle CAB`.
Answer» Correct Answer - `30^@`

Discussion

200.	Arcs cut by equal chords of a circle are:(A) equal(B) double(C) half(D) triple
Answer» Answer is (A) equal

Discussion

Explore topic-wise InterviewSolutions in .

Let O(0, 0) and (0, 1) be two fixed points, then the locus of a point P such that the perimeter of `DeltaAOP` is 4, isA. `8x^(2)-9y^(2)+9y=18`B. `9x^(2)-8y^(2)+8y=16`C. `9x^(2)+8y^(2)-8y=16`D. `8x^(2)+9y^(2)-9y=18`

Find the equation of a circle with: Centre at (2, -3) and radius 5

Find the equation of a circle with: centre at (-3, -2) and radius 6.

Find the equation of a circle with : centre at origin and radius 4.

P is a variable point on the line `L=0` . Tangents are drawn to the circles `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogram PQSR is completed. If `P -=(3,4)`, then the coordinates of S areA. `(-46//25,63//25)`B. `(-51//25,-68//25)`C. `(-46//25,68//25)`D. `(-68//25,51//25)`

P is a variable point on the line L=0. Tangents are drawn to the circle `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogran PQSR is completed. If P-=(6,8) then area of `DeltaQRS` is `(192)/(25)sqrtlamda` sq units. The value of `lamda` isA. 2B. 3C. 5D. 6

In the adjoining figure, AD is the diameter of the circle and `angleBCD=140^@`. Find the value of `angleADB`.

In the given figure, find `angle ABC`.

In the given figure , find the length of chord AB .

Find `angle AOB` in the given figure.

In the given figure, find `angle AOB`.

Find `angle A` in the given figure.

In the given figure, find `angleD` and `angleB`.

In the given figure , find `angleDCP`.

In the adjoining figure, AOB is the diameter of the circle. If `angleABP=45^@`, then find the value of `angle PQB`.

In the adjoining figure, `O` is the centre of the circle. If `anglePAO=15^@` and `anglePBO=30^@`, then find the value of `angleAOB`.

find the equation of circle whose centre `(2, 4 ) ` and radius 5

Find the centre and radius of each of the following circles : (i) ` (x - 3)^(2) + (y- 1) ^(2) = 9` (ii) ` (x - (1)/(2) ) ^(2) + ( y + (1)/(3) ) ^(2) = (1)/(16)` (iii) ` (x + 5) ^(2) + ( y- 3 ) ^(2) = 20` (iv)` x ^(2) + (y- 1 ) ^(2) = 2 `

Show that the points `(5, 5 ), (6, 4 ), (-2, 4) and (7, 1 )` are concyclic, i.e., all lie on the same circle. Find the equation, centre and radius of this circle.

Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____. (A) on the centre (B) inside the circle (C) outside the circle (D) on the circle

find the equation of circle whose centre ` (-3, - 2)` and radius 6

Show that the points `A(1, 0), B(2, -7), C(8, 1)` and `D(9,-6)` all lie on the same circle. Find the equation of this circle, its centre and radius.

centre at the origin and radius 4

centre ` (a, a)` and radius ` sqrt2`

Find the equation of the circle with centre : `(-a , b)`and radius `sqrt(a^2-b^2)`.

If the equation `p x^2+(2-q)x y+3y^2-6q x+30 y+6q=0`represents a circle, then find the values of `pa n dqdot`

Find the equation of the circle with: Centre `(a cosalpha, a s inalpha)`and radius `adot`

A circle is inscribed in an equilateral triangle of side `adot`The area of any square inscribed in this circle is ______.

If `x^2+y^2-2x+2a y+a+3=0`represents the real circle with nonzero radius, then find the values of`adot`A. `ain(-oo,-1)`B. `ain(-1,2)`C. `ain(2,oo)`D. `ain(-oo,-1)uu(2,oo)`

If `x^2+y^2-2x+2a y+a+3=0`represents the real circle with nonzero radius, then find the values of`adot`

In a cyclic `squareABCD` , twice the measure of `angleA` is thrice the measure of `angleC` . Find the measure of `angleC` .A. 36B. 72C. 90D. 108

A point `P`moves in such a way that the ratio of its distance from two coplanarpoints is always a fixed number `(!=1)`. Then, identify the locus of the point.

In the adjoining figure, `O` is the centre of the circle.If `angleBAC=40^@`, then find the value of `angleADC`.

ABCD is a cyclic trapezium in which, `AD| |BC` and `angleB=70^@`. Find its remaining angles,

In the adjoining figure , `O` is the center of the circle. If `angleBAD=30^(@)` , then find the values of x,y and z.

The angle made by the tangent and the radius made at the point of contact isA. `0^(@)`B. `45^(@)`C. `90^(@)`D. `75^(@)`

Find the image of the circle `x^2+y^2-2x+4y-4=0`in the line `2x-3y+5=0`

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that `/_A C P =/_Q C D`.

From the information given in the figure, find the measure of `/_AEC `. A. `42^(@)`B. `30^(@)`C. `36^(@)`D. `72^(@)`

A chord of 24 cm length is drawn in a circle of radius 13 cm. The distance of chord from center of circle is –(a) 12 cm(b) 5 cm(c) 6.5 cm(d) 12 cm

In given fig., O is center of the circle. Chord AB = CD and ∠OBA = 40°, then ∠COD will be :(A) 100°(B) 80°(C) 180°(D) 90°

In the adjoining figure, ABCD is a cyclic quadrilateral.If side BC is produced upto point E and `angleDAB=95^@`, then find the value of `angleDCE`.

In the adjoining figure, ABCD is a cyclic quadrilateral and `AB` is the diameter of the circle. If `angleAPC=120^@`, then find the value of `angle CAB`.

Arcs cut by equal chords of a circle are:(A) equal(B) double(C) half(D) triple