Class 11 - Mathematics
Conic Sections - Exercise 11.1
Top Block 1
Question 1:
Find the equation of the circle with centre (0, 2) and radius 2.
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
=> x2 + y2 + 4 – 4y = 4
=> x2 + y2 – 4y = 0
Question 2:
Find the equation of the circle with centre (–2, 3) and radius 4.
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is
(x + 2)2 + (y – 3)2 = (4)2
=> x2 + 4x + 4 + y2 – 6y + 9 = 16
=> x2 + y2 + 4x – 6y – 3 = 0
Question 3:
Find the equation of the circle with centre (1/2, 1/4) and radius 1/12.
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (1/2, 1/4) and radius (r) = 1/12
Therefore, the equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = (1/12)2
=> x2 – x + 1/4 + y2 – y/2 + 1/16 = 1/144
=> x2 – x + 1/4 + y2 – y/2 + 1/16 – 1/144 = 0
=> (144x2 – 144x + 36 + 144y2 – 72y + 9 – 1)/144 = 0
=> 144x2 – 144x + 36 + 144y2 – 72y + 9 – 1 = 0
=> 144x2 – 144x + 144y2 – 72y + 44 = 0
=> 4(36x2 – 36x + 36y2 – 18y + 11) = 0
=> 36x2 – 36x + 36y2 – 18y + 11 = 0
=> 36x2 + 36y2 – 36x – 18y + 11 = 0
Question 4:
Find the equation of the circle with centre (1, 1) and radius √2.
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (1, 1) and radius (r) = √2.
Therefore, the equation of the circle is
(x – 1)2 + (y – 1)2 = (√2)2
=> x2 – 2x + 1 + y2 – 2y + 1 = 2
=> x2 + y2 –2x – 2y = 0
Question 5:
Find the equation of the circle with centre (–a, –b) and radius √(a2 − b2).
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–a, –b) and radius (r) = √(a2 − b2)
Therefore, the equation of the circle is
(x + a)2 + (y + b)2 = {√(a2 − b2)}2
=> x2 + 2ax + a2 + y2 + 2by + b2 = a2 − b2
=> x2 + y2 + 2ax + 2by + 2b2 = 0
Question 6:
Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36
Answer:
The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.
(x + 5)2 + (y – 3)2 = 36
=> {x – (–5)}2 + (y – 3)2 = 62
Which is of the form (x – h)2 + (y – k)2 = r2
Where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3) and its radius is 6.
Question 7:
Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0
Answer:
The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0
=> x2 + y2 – 4x – 8y – 45 = 0
=> (x2 – 4x) + (y2 – 8y) = 45
=> {x2 – 2 * x * 2 + 22} + {y2 – 2 * y * 4+ 42} – 4 –16 = 45
=> (x – 2)2 + (y –4)2 = 65
=> (x – 2)2 + (y –4)2 = (√65)2
Which is of the form (x – h)2 + (y – k)2 = r2
Where h = 2, k = 4 and r = √65
Thus, the centre of the given circle is (2, 4), while its radius is √65
Question 8:
Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0
Answer:
The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0
=> x2 + y2 – 8x + 10y – 12 = 0
=> (x2 – 8x) + (y2 + 10y) = 12
=> {x2 – 2 * x * 4 + 42} + {y2 + 2 * y * 5 + 52} – 16 –25 = 12
=> (x – 4)2 + (y + 5)2 = 53 => (x – 4)2 + (y + 5)2 = (√53)2
Which is of the form (x – h)2 + (y – k)2 = r2
Where h = 4, k = -5 and r = √53
Thus, the centre of the given circle is (4, -5), while its radius is √53.
Question 9:
Find the centre and radius of the circle 2x2 + 2y2 – x = 0
Answer:
The equation of the given circle is 2x2 + 2y2 – x = 0
=> 2x2 + 2y2 – x = 0
=> 2x2 – x + 2y2 = 0
=> 2(x2 – x/2 + y2) = 0
=> x2 – x/2 + y2 = 0
=> x2 – 2 * x * (1/4) + (1/4)2 + y2 – (1/4)2 = 0
=> (x – 1/4)2 + (y – 0)2 = (1/4)2
which is of the form (x – h)2 + (y – k)2 = r2
where h = 1/4, k = 0 and r = 1/4
Thus, the centre of the given circle is (1/4, 0), while its radius is 1/4.
Question 10:
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 …………………. 1
(6 – h)2 + (5 – k)2 = r2 …………………. 2
Since the centre (h, k) of the circle lies on line 4x + y = 16,
4h + k = 16 …………………3
From equations (1) and (2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
=> 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
=> 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
=> 4h + 8k = 44
=> h + 2k = 11 ………………4
On solving equations 3 and 4, we obtain h = 3 and k = 4.
On substituting the values of h and k in equation 1, we obtain
(4 – 3)2 + (1 – 4)2 = r2
=> (1)2 + (– 3)2 = r2
=> 1 + 9 = r2
=> r2 = 10
=> r = √10
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = (√10)2
=> x2 – 6x + 9 + y2 – 8y + 16 = 10
=> x2 + y2 – 6x – 8y + 15 = 0
Question 11:
Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2
Since the circle passes through points (2, 3) and (–1, 1),
(2 – h)2 + (3 – k)2 = r2 ……………….1
(–1 – h)2 + (1 – k)2 = r2 ……………..2
Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,
So, h – 3k = 11 …………………3
From equations 1 and 2, we obtain
(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2
=> 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2
=> 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
=> 6h + 4k = 11 …………….4
On solving equations (3) and (4), we obtain h = 7/2and k = −5/2
On substituting the values of h and k in equation 1, we obtain
(2 – 7/2)2 + (3 + 5/2)2 = r2
=> (-3/2)2 + (11/2)2 = r2
=> 9/4 + 121/4 = r2
=> r2 = 130/4
Thus, the equation of the required circle is
(x – 7/2)2 + (y + 5/2)2 = 130/4
=> {(2x – 7)/2}2 + {(2y + 5)/2}2 = 130/4
=> 4x2 – 28x + 49 + 4y2 + 20y + 25 = 130
=> 4x2 – 28x + 49 + 4y2 + 20y + 25 – 130 = 0
=> 4x2 – 28x + 4y2 + 20y – 56 = 0
=> 4(x2 + y2 – 7x + 5y – 14) = 0
=> x2 + y2 – 7x + 5y – 14 = 0
Mddle block 1
Question 12:
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)2 + y2 = 25
It is given that the circle passes through point (2, 3).
So, (2 – h)2 + 32 = 25
=> (2 – h)2 + 9 = 25
=> (2 – h)2 = 25 – 9
=> (2 – h)2 = 16
=> 2 – h = ±√16
=> 2 – h = ±4
If 2 – h = 4, then h = -2
If 2 – h = -4, then h = 6
When h = –2, the equation of the circle becomes
(x + 2)2 + y2 = 25
=> x2 + 4x + 4 + y2 = 25
=> x2 + y2 + 4x – 21 = 0
When h = 6, the equation of the circle becomes
(x – 6)2 + y2 = 25
=> x2 – 12x + 36 + y2 = 25
=> x2 + y2 – 12x + 11 = 0
Question 13:
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
=> h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the
circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2 …………………… (1)
(0 – h)2 + (b – k)2 = h2 + k2 …………………… (2)
From equation (1), we obtain
a2 – 2ah + h2 + k2 = h2 + k2
=> a2 – 2ah = 0
=> a(a – 2h) = 0
=> a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0
=> h =a/2
From equation (2), we obtain
h2 + b2 – 2bk + k2 = h2 + k2
=> b2 – 2bk = 0
=> b(b – 2k) = 0
=> b = 0 or (b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0
=> k =b/2
Thus, the equation of the required circle is
(x – a/2)2 + (y – b/2)2 = (a/2)2 + (b/2)2
=> {(2x – a)/2}2 + {(2y – b)/2}2 = a2/4 + b2/4
=> (4x2 – 4ax + a2)/4 + (4y2 – 4by + b2)/4 = (a2 + b2)/4
=> 4x2 – 4ax + a2 + 4y2 – 4by + b2 = a2 + b2
=> 4x2 + 4y2 – 4ax – 4by = 0
=> 4(x2 + y2 – ax – by) = 0
=> x2 + y2 – ax – by = 0
Question 14:
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Answer:
The centre of the circle is given as (h, k) = (2, 2)
Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between
the points (2, 2) and (4, 5).
So, r = √{(2 – 4)2 + (2 – 5)2} = √{(-2)2 + (-3)2} = √(4 + 9) = √13
Thus, the equation of the circle is
(x – h)2 + (y – k)2 = r2
=> (x – 2)2 + (y – 2)2 = (√13)2
=> x2 – 4x + 4 + y2 – 4y + 4 = 13
=> x2 + y2 – 4x – 4y – 5 = 0
Question 15:
Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Answer:
The equation of the given circle is x2 + y2 = 25
=> x2 + y2 = 25
=> (x – 0)2 + (y – 0)2 = 52
which is of the form (x – h)2 + (y – k)2 = r2
where h = 0, k = 0, and r = 5.
So, Centre = (0, 0) and radius = 5
Now, distance between point (–2.5, 3.5) and centre (0, 0)
= √{(-2.5 – 0)2 + (3.5 – 0)2}
= √{(-2.5)2 + (3.5)2}
= √(6.25 + 12.25)
= √18.5
= 4.3 < 5
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the
radius of the circle,
So, the point (-2.5, 3.5) lies inside the circle.
Bottom Block 3
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