Very Short Answer on Circle Class 11 OP Malhotra Exe-17E ISC Maths Solutions Ch-17. In this article you would learn to solve all type 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 Very Short Answer ISC Math Solutions Ch-17
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-17 | Circle |
| Writer | OP Malhotra |
| Exe-17(E) | Very Short Answer Type Questions. |
Very Short Answer on Circle
OP Malhotra ISC Class 11 Maths Solutions
Que-1: Find the centre and radius of the circle 2x² + 2y² − x = 0
Sol: Divide the equation by 2:
x² + y² − x/2 = 0
Complete the square:
x² − x/2 = (x − 1/4)² − 1/16
So equation becomes:
(x − 1/4)² + y² = 1/16
Centre = (1/4, 0), Radius = 1/4
Que-2: Find the equation of the circle with centre (−a, −b) and radius √(a² − b²)
Sol: Standard form: (x − h)² + (y − k)² = r²
Here h = −a, k = −b
⇒ (x + a)² + (y + b)² = a² + b²
x² + a² + 2ax + y² + b² + 2by = a² – b²
x² + y² + 2ax + 2by = a² – b² – a² – b²
x² + y² + 2ax + 2by = 2b²
x² + y² + 2ax + 2by – 2b² = 0
Que-3: Find the equation of the circle drawn on the line joining (−1, 2) and (3, −4) as diameter
Sol: Midpoint (centre) = ( (−1+3)/2 , (2−4)/2 ) = (1, −1)
Radius = half distance:
= √[(4)² + (−6)²] / 2 = √52 / 2 = √13
Equation: (x − 1)² + (y + 1)² = 13
x² + 1 – 2x + y² + 1 + 2y = 13
x² + y² – 2x + 2y = 13 – 2
x² + y² – 2x + 2y – 11 = 0.
Que-4: Find the equation of the circle concentric with x² + y² − 8x + 6y − 5 = 0 and passing through (−2, −7)
Sol: Given circle centre = (4, −3)
Required circle has same centre: (4, −3)
Radius = distance from (4, −3) to (−2, −7):
= √[(−6)² + (−4)²] = √52
Equation: (x − 4)² + (y + 3)² = 52
x² + 16 – 8x + y² + 9 + 6y = 52
x² + y² – 8x + 6y + 25 = 52
x² + y² – 8x + 6y + 25 – 52 = 0
x² + y² – 8x + 6y – 27 = 0
Que-5: Find the equation of a circle with centre (2, 2) and passing through (4, 5)
Sol: Radius = distance between (2,2) and (4,5)
= √[(2)² + (3)²] = √13
Equation: (x − 2)² + (y − 2)² = 13
x² + 4 – 4x + y² + 4 – 4y = 13
x² + y² – 4x – 4y + 8 = 13
x² + y² – 4x – 4y + 8 – 13
x² + y² – 4x – 4y – 5
Que-6: Find whether the point (−1, 3) lies inside, outside or on the circle x² + y² = 16
Sol: Substitute point:
(−1)² + 3² = 1 + 9 = 10
Since 10 < 16
⇒ point lies inside the circle.
Que-7: Find the equation of the circle which touches both axes in the first quadrant and whose radius is a
Sol: Centre = (a, a) because it touches both axes
Equation: (x − a)² + (y − a)² = a²
x² + a² – 2ax + y² + a² – 2ay = a²
x² + y² – 2ax – 2ay + 2a² – a² = 0
x² + y² – 2ax – 2ay + a² = 0
Que-8: Find the equation of the circle which touches x-axis and whose centre is (1, 2)
Sol: Radius = distance from centre to x-axis = 2
Equation: (x − 1)² + (y − 2)² = 4
x² + 1 – 2x + y² + 4 – 4y = 4
x² + y² – 2x – 4y + 5 – 4 = 0
x² + y² – 2x – 4y + 1 = 0
Que-9: If the line y = √3 x + k touches x² + y² = 16, find k
Sol: Circle centre = (0,0), radius = 4
Distance of line from centre = radius:
|k| / √(1 + (√3)²) = 4
⇒ |k| / 2 = 4
⇒ |k| = 8
k = ±8
–: End Circle Class 11 OP Malhotra Exe-17E ISC Maths Ch-17 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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