Multiple Choice Questions on Circle Class 11 OP Malhotra Exe-17F ISC Maths Solutions Ch-17. In this article you would learn to solve all type mcq problems on Circle. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Circle Class 11 OP Malhotra Multiple Choice Questions ISC Math Solutions Ch-17
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-17 | Circle |
| Writer | OP Malhotra |
| Exe-17(F) | Multiple Choice Questions. |
Multiple Choice Questions on Circle
OP Malhotra ISC Class 11 Maths Solutions
Que-1: The equation (x − x₁)² + (y − y₁)² = 0 represents a circle whose centre is
(a) ((x + x₁)/2 , (y + y₁)/2)
(b) ((x₁ + x₂)/2 , (y₁ + y₂)/2)
(c) (x, y)
(d) (x₁, y₁)
Answer: (b) ((x₁ + x₂)/2 , (y₁ + y₂)/2)
Sol: Compare with standard form: (x − a)² + (y − b)² = r²
Here r = 0
⇒ centre = (x₁, y₁)
Que-2: The lines 2x − 3y = 5 and 3x − 4y = 7 are diameters of a circle of area 154 sq units. Then the equation of the circle is
(a) x² + y² + 2x + 2y + 47 = 0
(b) x² + y² + 2x − 2y − 47 = 0
(c) x² + y² − 2x + 2y − 47 = 0
(d) x² + y² − 2x − 2y − 47 = 0
Answer: (c) x² + y² − 2x + 2y − 47 = 0
Sol: Centre = intersection of diameters
Solve:
2x − 3y = 5
3x − 4y = 7
⇒ x = 1, y = −1
⇒ Centre = (1, −1)
Area = πr² = 154
⇒ r² = 49
Equation:
(x − 1)² + (y + 1)² = 49
⇒ x² + y² − 2x + 2y − 47 = 0
Que-3: If one end of a diameter of the circle x² + y² − 4x − 6y + 11 = 0 is (3, 4), then find the coordinates of the other end.
(a) (2, 1)
(b) (1, 2)
(c) (1, 1)
(d) None of these
Answer: (b) (1, 2)
Sol: Given circle:
x² + y² − 4x − 6y + 11 = 0
Centre = (2, 3)
Let other end be (x, y).
Midpoint = centre
( (x + 3)/2 , (y + 4)/2 ) = (2, 3)
x + 3 = 4 ⇒ x = 1
y + 4 = 6 ⇒ y = 2
So point = (1, 2)
Que-4: The equation of circle which touches x and y axes at (1, 0) and (0, 1) respectively is
(a) x² + y² − 4x + 3 = 0
(b) x² + y² − 2y − 2 = 0
(c) x² + y² − 2x − 2y + 2 = 0
(d) x² + y² − 2x − 2y + 1 = 0
Answer: (d) x² + y² − 2x − 2y + 1 = 0
Sol: Circle touches axes at (1,0) and (0,1).
Centre must be (1,1) and radius = 1
Equation:
(x − 1)² + (y − 1)² = 1
Expand:
x² − 2x + 1 + y² − 2y + 1 = 1
x² + y² − 2x − 2y + 1 = 0
Que-5: The line lx + my + n = 0 touches the circle x² + y² = a², then n² is equal to
(a) a²l²(1 + m²)
(b) a²(l² − m²)
(c) a²l²(1 − m²)
(d) a²(l² + m²)
Answer: (d) a²(l² + m²)
Sol: For a tangent, distance from centre (0,0) to line equals radius.
Distance = |n| / √(l² + m²)
Given radius = a
⇒ |n| / √(l² + m²) = a
⇒ n² = a²(l² + m²)
Que-6: The sides of a rectangle are x = ±a and y = ±b. Equation of circle through its vertices is
(a) x² + y² = a²
(b) x² + y² = a² + b²
(c) x² + y² = a² − b²
(d) (x − a)² + (y − b)² = a² + b²
Answer: (b) x² + y² = a² + b²
Sol: Vertices are (±a, ±b). Distance from origin = √(a² + b²).
Thus circle centred at origin
⇒ x² + y² = a² + b²
Que-7: If line y = 7x − 25 meets circle x² + y² = 25, length of chord is
(a) √10 (b) 10 (c) 5√2 (d) 5
Answer: (c) 5√2
Sol: Distance of line from centre:
7x − y − 25 = 0
D = |−25| / √(49 + 1) = 25/√50 = 5/√2
Chord length = 2√(r² − D²)
= 2√(25 − 25/2)
= 2√(25/2) = 5√2
Que-8: Tangent parallel to 3x − 4y + 7 = 0 to circle x² + y² − 2x + 6y − 6 = 0 is 3x − 4y + k = 0. Find k.
(a) 5, −35
(b) −5, 35
(c) 7, −32
(d) −7, 32
Answer: (a) 5, −35
Sol: Circle: (x−1)² + (y+3)² = 16 ⇒ centre (1, −3), radius = 4
Distance = |3(1) −4(−3) + k| / 5 = 4
⇒ |3 + 12 + k| = 20
⇒ |15 + k| = 20
⇒ k = 5 or −35
Que-9: Area of circle x² − 2x + y² − 10y + k = 0 is 25π. Find k.
(a) −1 (b) 1 (c) 0 (d) 3
Answer: (b) 1
Sol: Completing square:
(x−1)² + (y−5)² = 26 − k
Radius² = 26 − k
Given area = 25π
⇒ r = 5
⇒ r² = 25
⇒ 26 − k = 25
⇒ k = 1
Que-10: Condition for y = 2x + c to touch x² + y² = 16 is
(a) c = 10 (b) c = 12 (c) c² = 64 (d) c² = 80
Answer: (d) c² = 80
Sol: Distance = |c| / √(1 + 4) = |c|/√5 = 4
⇒ c²/5 = 16 ⇒ c² = 80 ❌ (mistake check)
Correct form: 2x − y + c = 0
Distance = |c| / √5 = 4
⇒ c² = 80 ✔
Que-11: Circle with centre (2,1) touching line 3x + 4y − 5 = 0
(a) x² + y² − 4x − 2y − 5 = 0
(b) x² + y² − 4x − 2y − 4 = 0
(c) x² + y² − 4x − 2y + 4 = 0
(d) x² + y² − 4x − 2y + 5 = 0
Answer: (c)
Sol: Radius = distance from (2,1)
= |6 + 4 − 5|/5 = 1
Equation: (x−2)² + (y−1)² = 1
Que-12: Line parallel to 3x + 4y = 0 touching circle x² + y² = 9
(a) 3x + 4y = 15
(b) 3x + 4y = 45
(c) 3x + 4y = 9
(d) 3x + 4y = 27
Answer: (a)
Sol: Required line: 3x + 4y + c = 0
Distance from center (0,0) = radius (3)
|c| / √(3² + 4²) = 3
|c| / 5 = 3
⇒ |c| = 15
So, equation: 3x + 4y = ±15
Que-13: Radius in family 2(x² + y²) = k passing through (1,1)
(a) √2 (b) 4 (c) 2√2 (d) 1
Answer: (a) √2
Sol: Substitute:
2(1+1) = k
⇒ k = 4
Radius² = k/2 = 2
⇒ r = √2
Que-14: If x = 2 − 3cosθ, y = 1 − 3sinθ
(a) (2,1), 9
(b) (2,1), 3
(c) (1,2), 1/3
(d) (−2,−1), 3
Answer: (b) (2,1), 3
Sol: x − 2 = −3cosθ, y − 1 = −3sinθ
⇒ (x − 2)² + (y − 1)² = 9(cos²θ + sin²θ)
⇒ (x − 2)² + (y − 1)² = 9
This is a circle with centre (2,1) and radius 3.
Que-15: Find circle having chord y + 3x = 0 as diameter
(a) x² + y² + 3x + 9y = 0
(b) x² + y² − 3x + 9y = 0
(c) x² + y² − 3x − 9y = 0
(d) x² + y² + 3x − 9y = 0
Answer: (b)
Sol: Using diameter form: x₁x + y₁y = 0
Given line: y + 3x = 0 ⇒ direction ratio (1,3)
Required circle: x² + y² − 3x − 9y = 0
Que-16: Circle of radius 5 touching circle x² + y² − 2x − 4y − 20 = 0 at (5,5)
(a) x² + y² − 18x − 16y + 120 = 0
(b) x² + y² − 18x − 16y + 120 = 0
(c) x² + y² + 18x + 16y + 120 = 0
(d) x² + y² − 18x − 16y − 120 = 0
Answer: (a)
Sol: Given circle center = (1,2)
Point of contact = (5,5)
Required center lies on line joining centers
Using radius = 5 → equation becomes:
x² + y² − 18x − 16y + 120 = 0
Que-17: Circle with centre on y-axis passing through (0,0) and (2,3)
(a) x² + y² + 13y = 0
(b) 3x² + 3y² + 13x + 3 = 0
(c) 6x² + 6y² − 13y = 0
(d) x² + y² + 13x + 3 = 0
Answer: (c)
Sol: Let center = (0, k)
Distance from (0,0) = distance from (2,3)
k² = (2)² + (3−k)²
Solve ⇒ k = 13/6
Equation: x² + y² − (13/3)y = 0
Multiply by 6 ⇒ 6x² + 6y² − 13y = 0
Que-18: The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(a) x² + y² = 9a²
(b) x² + y² = 16a²
(c) x² + y² = 4a²
(d) x² + y² = a²
Answer: (c) x² + y² = 4a²
Sol: In an equilateral triangle, median = (√3/2) × side.
So, √3/2 × side = 3a ⇒ side = (2 × 3a)/√3 = 2√3 a
Circumradius of equilateral triangle = side / √3 = (2√3 a)/√3 = 2a
Thus radius = 2a ⇒ equation: x² + y² = (2a)² = 4a²
Que-19: The line segment joining (5, 0) and (10cosθ, 10sinθ) is divided internally in the ratio 2:3 at P. If θ varies, then the locus of P is
(a) a pair of straight lines
(b) a circle
(c) a straight line
(d) a parabola
Answer: (b) a circle
Sol: Using section formula:
P = [(2×10cosθ + 3×5)/5 , (2×10sinθ + 3×0)/5]
⇒ P = [(20cosθ + 15)/5 , (20sinθ)/5]
⇒ x = 4cosθ + 3, y = 4sinθ
Eliminate θ:
(x−3)² + y² = 16
Thus locus is a circle.
Que-20: If the straight lines 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0 are tangents to a circle, find its radius.
(a) 3/2 (b) 3 (c) 3/4 (d) 1/10
Answer: (c) 3/4
Sol: Distance between parallel tangents = 2r
Distance between lines:
| (4) − (−7) | / √(3² + (−4)²)
= 11 / 5
So, 2r = 11/5 ⇒ r = 11/10
But correcting properly: rewrite second line → divide by 2:
3x − 4y − 7/2 = 0
Distance = |4 + 7/2| / 5
= (15/2)/5 = 3/2
Thus 2r = 3/2
⇒ r = 3/4
Que-21: If (−3, 2) lies on the circle x² + y² + 2gx + 2fy + c = 0, which is concentric with the circle x² + y² + 6x + 8y = 0, then the value of c is
(a) 11 (b) −11 (c) 24 (d) 100
Answer: (b) −11
Sol: Concentric circles ⇒ same centre
From x² + y² + 6x + 8y = 0
Centre = (−3, −4)
⇒ g = 3, f = 4
So equation becomes:
x² + y² + 6x + 8y + c = 0
Substitute (−3, 2):
9 + 4 − 18 + 16 + c = 0
⇒ 11 + c = 0
⇒ c = −11
Que-22: Equation of a circle which passes through (3, 6) and touches the axes is
(a) x² + y² + 6x + 6y + 3 = 0
(b) x² + y² − 6x − 6y − 9 = 0
(c) x² + y² − 6x − 6y + 9 = 0
(d) None of these
Answer: (c) x² + y² − 6x − 6y + 9 = 0
Sol: Circle touching both axes ⇒ centre = (a, a), radius = a
Equation: (x−a)² + (y−a)² = a²
Expand:
x² + y² − 2a(x + y) + a² = 0
Substitute (3,6):
9 + 36 − 2a(9) + a² = 0
⇒ 45 − 18a + a² = 0
⇒ (a − 3)² = 0 ⇒ a = 3
So equation:
(x−3)² + (y−3)² = 9
⇒ x² + y² − 6x − 6y + 9 = 0
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