RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - The Circle

 Page 1


24. The Circle
Exercise 24.1
1 A. Question
Find the equation of the circle with:
Centre ( - 2, 3) and radius 4.
Answer
(i) Given that we need to find the equation of the circle with centre ( - 2, 3) and radius 4.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - ( - 2))
2
 + (y - 3)
2
 = 4
2
? (x + 2)
2
 + (y - 3)
2
 = 16
? x
2
 + 4x + 4 + y
2
 - 6y + 9 = 16
? x
2
 + y
2
 + 4x - 6y - 3 = 0
?The equation of the circle is x
2
 + y
2
 + 4x - 6y - 3 = 0.
1 B. Question
Find the equation of the circle with:
Centre (a, b) and radius .
Answer
Given that we need to find the equation of the circle with centre (a, b) and radius .
Page 2


24. The Circle
Exercise 24.1
1 A. Question
Find the equation of the circle with:
Centre ( - 2, 3) and radius 4.
Answer
(i) Given that we need to find the equation of the circle with centre ( - 2, 3) and radius 4.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - ( - 2))
2
 + (y - 3)
2
 = 4
2
? (x + 2)
2
 + (y - 3)
2
 = 16
? x
2
 + 4x + 4 + y
2
 - 6y + 9 = 16
? x
2
 + y
2
 + 4x - 6y - 3 = 0
?The equation of the circle is x
2
 + y
2
 + 4x - 6y - 3 = 0.
1 B. Question
Find the equation of the circle with:
Centre (a, b) and radius .
Answer
Given that we need to find the equation of the circle with centre (a, b) and radius .
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - b)
2
 = 
? x
2
 - 2ax + a
2
 + y
2
 - 2by + b
2
 = a
2
 + b
2
? x
2
 + y
2
 - 2ax - 2by = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2by = 0.
1 C. Question
Find the equation of the circle with:
Centre (0, - 1) and radius 1.
Answer
Given that we need to find the equation of the circle with centre (0, - 1) and radius 1.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - 0)
2
 + (y - ( - 1))
2
 = 1
2
? (x - 0)
2
 + (y + 1)
2
 = 1
? x
2
 + y
2
 + 2y + 1 = 1
? x
2
 + y
2
 + 2y = 0
Page 3


24. The Circle
Exercise 24.1
1 A. Question
Find the equation of the circle with:
Centre ( - 2, 3) and radius 4.
Answer
(i) Given that we need to find the equation of the circle with centre ( - 2, 3) and radius 4.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - ( - 2))
2
 + (y - 3)
2
 = 4
2
? (x + 2)
2
 + (y - 3)
2
 = 16
? x
2
 + 4x + 4 + y
2
 - 6y + 9 = 16
? x
2
 + y
2
 + 4x - 6y - 3 = 0
?The equation of the circle is x
2
 + y
2
 + 4x - 6y - 3 = 0.
1 B. Question
Find the equation of the circle with:
Centre (a, b) and radius .
Answer
Given that we need to find the equation of the circle with centre (a, b) and radius .
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - b)
2
 = 
? x
2
 - 2ax + a
2
 + y
2
 - 2by + b
2
 = a
2
 + b
2
? x
2
 + y
2
 - 2ax - 2by = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2by = 0.
1 C. Question
Find the equation of the circle with:
Centre (0, - 1) and radius 1.
Answer
Given that we need to find the equation of the circle with centre (0, - 1) and radius 1.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - 0)
2
 + (y - ( - 1))
2
 = 1
2
? (x - 0)
2
 + (y + 1)
2
 = 1
? x
2
 + y
2
 + 2y + 1 = 1
? x
2
 + y
2
 + 2y = 0
?The equation of the circle is x
2
 + y
2
 + 2y = 0.
1 D. Question
Find the equation of the circle with:
Centre (a cos , a sin ) and radius a.
Answer
Given that we need to find the equation of the circle with centre (acos a, asina) and radius a.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - acosa)
2
 + (y - asina)
2
 = a
2
? x
2
 - (2acosa)x + a
2
cos
2
a + y
2
 - (2asina)y + a
2
sin
2
a = a
2
We know that sin
2
? + cos
2
? = 1
? x
2
 - (2acosa)x + y
2
 - 2asinay + a
2
 = a
2
? x
2
 + y
2
 - (2acosa)x - (2asina)y = 0
?The equation of the circle is x
2
 + y
2
 - (2acosa)x - (2asina)y = 0.
1 E. Question
Find the equation of the circle with:
Centre (a, a) and radius v2 a.
Answer
Given that we need to find the equation of the circle with centre (a, a) and radius v2a.
Page 4


24. The Circle
Exercise 24.1
1 A. Question
Find the equation of the circle with:
Centre ( - 2, 3) and radius 4.
Answer
(i) Given that we need to find the equation of the circle with centre ( - 2, 3) and radius 4.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - ( - 2))
2
 + (y - 3)
2
 = 4
2
? (x + 2)
2
 + (y - 3)
2
 = 16
? x
2
 + 4x + 4 + y
2
 - 6y + 9 = 16
? x
2
 + y
2
 + 4x - 6y - 3 = 0
?The equation of the circle is x
2
 + y
2
 + 4x - 6y - 3 = 0.
1 B. Question
Find the equation of the circle with:
Centre (a, b) and radius .
Answer
Given that we need to find the equation of the circle with centre (a, b) and radius .
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - b)
2
 = 
? x
2
 - 2ax + a
2
 + y
2
 - 2by + b
2
 = a
2
 + b
2
? x
2
 + y
2
 - 2ax - 2by = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2by = 0.
1 C. Question
Find the equation of the circle with:
Centre (0, - 1) and radius 1.
Answer
Given that we need to find the equation of the circle with centre (0, - 1) and radius 1.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - 0)
2
 + (y - ( - 1))
2
 = 1
2
? (x - 0)
2
 + (y + 1)
2
 = 1
? x
2
 + y
2
 + 2y + 1 = 1
? x
2
 + y
2
 + 2y = 0
?The equation of the circle is x
2
 + y
2
 + 2y = 0.
1 D. Question
Find the equation of the circle with:
Centre (a cos , a sin ) and radius a.
Answer
Given that we need to find the equation of the circle with centre (acos a, asina) and radius a.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - acosa)
2
 + (y - asina)
2
 = a
2
? x
2
 - (2acosa)x + a
2
cos
2
a + y
2
 - (2asina)y + a
2
sin
2
a = a
2
We know that sin
2
? + cos
2
? = 1
? x
2
 - (2acosa)x + y
2
 - 2asinay + a
2
 = a
2
? x
2
 + y
2
 - (2acosa)x - (2asina)y = 0
?The equation of the circle is x
2
 + y
2
 - (2acosa)x - (2asina)y = 0.
1 E. Question
Find the equation of the circle with:
Centre (a, a) and radius v2 a.
Answer
Given that we need to find the equation of the circle with centre (a, a) and radius v2a.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - a)
2
 = ( a)
2
? x
2
 - 2ax + a
2
 + y
2
 - 2ay + a
2
 = 2a
2
? x
2
 + y
2
 - 2ax - 2ay = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2ay = 0.
2 A. Question
Find the centre and radius of each of the following circles:
(x - 1)
2
 + y
2
 = 4
Answer
(i) Given that we need to find the centre and radius of the given circle (x - 1)
2
 + y
2
 = 4.
We know that the standard equation of a circle with centre (a, b) and having radius ‘r’ is given by:
? (x - a)
2
 + (y - b)
2
 = r
2
 - - - - - (1)
Let us convert given circle’s equation into the standard form.
? (x - 1)
2
 + y
2
 = 4
? (x - 1)
2
 + (y - 0)
2
 = 2
2
 ..... - (2)
Comparing (2) with (1), we get
? Centre = (1, 0) and radius = 2
Page 5


24. The Circle
Exercise 24.1
1 A. Question
Find the equation of the circle with:
Centre ( - 2, 3) and radius 4.
Answer
(i) Given that we need to find the equation of the circle with centre ( - 2, 3) and radius 4.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - ( - 2))
2
 + (y - 3)
2
 = 4
2
? (x + 2)
2
 + (y - 3)
2
 = 16
? x
2
 + 4x + 4 + y
2
 - 6y + 9 = 16
? x
2
 + y
2
 + 4x - 6y - 3 = 0
?The equation of the circle is x
2
 + y
2
 + 4x - 6y - 3 = 0.
1 B. Question
Find the equation of the circle with:
Centre (a, b) and radius .
Answer
Given that we need to find the equation of the circle with centre (a, b) and radius .
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - b)
2
 = 
? x
2
 - 2ax + a
2
 + y
2
 - 2by + b
2
 = a
2
 + b
2
? x
2
 + y
2
 - 2ax - 2by = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2by = 0.
1 C. Question
Find the equation of the circle with:
Centre (0, - 1) and radius 1.
Answer
Given that we need to find the equation of the circle with centre (0, - 1) and radius 1.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - 0)
2
 + (y - ( - 1))
2
 = 1
2
? (x - 0)
2
 + (y + 1)
2
 = 1
? x
2
 + y
2
 + 2y + 1 = 1
? x
2
 + y
2
 + 2y = 0
?The equation of the circle is x
2
 + y
2
 + 2y = 0.
1 D. Question
Find the equation of the circle with:
Centre (a cos , a sin ) and radius a.
Answer
Given that we need to find the equation of the circle with centre (acos a, asina) and radius a.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - acosa)
2
 + (y - asina)
2
 = a
2
? x
2
 - (2acosa)x + a
2
cos
2
a + y
2
 - (2asina)y + a
2
sin
2
a = a
2
We know that sin
2
? + cos
2
? = 1
? x
2
 - (2acosa)x + y
2
 - 2asinay + a
2
 = a
2
? x
2
 + y
2
 - (2acosa)x - (2asina)y = 0
?The equation of the circle is x
2
 + y
2
 - (2acosa)x - (2asina)y = 0.
1 E. Question
Find the equation of the circle with:
Centre (a, a) and radius v2 a.
Answer
Given that we need to find the equation of the circle with centre (a, a) and radius v2a.
We know that the equation of the circle with centre (p, q) and radius ‘r’ is given by:
? (x - p)
2
 + (y - q)
2
 = r
2
Now we substitute the corresponding values in the equation:
? (x - a)
2
 + (y - a)
2
 = ( a)
2
? x
2
 - 2ax + a
2
 + y
2
 - 2ay + a
2
 = 2a
2
? x
2
 + y
2
 - 2ax - 2ay = 0
?The equation of the circle is x
2
 + y
2
 - 2ax - 2ay = 0.
2 A. Question
Find the centre and radius of each of the following circles:
(x - 1)
2
 + y
2
 = 4
Answer
(i) Given that we need to find the centre and radius of the given circle (x - 1)
2
 + y
2
 = 4.
We know that the standard equation of a circle with centre (a, b) and having radius ‘r’ is given by:
? (x - a)
2
 + (y - b)
2
 = r
2
 - - - - - (1)
Let us convert given circle’s equation into the standard form.
? (x - 1)
2
 + y
2
 = 4
? (x - 1)
2
 + (y - 0)
2
 = 2
2
 ..... - (2)
Comparing (2) with (1), we get
? Centre = (1, 0) and radius = 2
?The centre and radius of the circle is (1, 0) and 2.
2 B. Question
Find the centre and radius of each of the following circles:
(x + 5)
2
 + (y + 1)
2
 = 9
Answer
Given that we need to find the centre and radius of the given circle (x + 5)
2
 + (y + 1)
2
 = 9.
We know that the standard equation of a circle with centre (a, b) and having radius ‘r’ is given by:
? (x - a)
2
 + (y - b)
2
 = r
2
 - - - - - (1)
Let us convert given circle’s equation into the standard form.
? (x + 5)
2
 + (y + 1)
2
 = 9
? (x - ( - 5))
2
 + (y - ( - 1))
2
 = 3
2
 - - - - (2)
Comparing (2) with (1), we get
? Centre = ( - 5, - 1) and radius = 3
?The centre and radius of the circle is ( - 5, - 1) and 3.
2 C. Question
Find the centre and radius of each of the following circles:
x
2
 + y
2
 - 4x + 6y = 5
Answer
Given that we need to find the centre and radius of the given circle x
2
 + y
2
 - 4x + 6y = 5.