Chapter 11 Conic Sections Ex 11.1 Solutions
Question - 1 : - Find the equation of the circle with centre (0, 2) and radius 2
Answer - 1 : -
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 ofthe circle is
(x – 0)2 + (y –2)2 = 22
x2 + y2 + 4 – 4 y =4
x2 + y2 – 4y =0
Question - 2 : - Find the equation of the circle with centre (–2, 3) and radius 4
Answer - 2 : -
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 ofthe 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 
and radius 
Answer - 3 : -
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)= and radius (r) =.
Therefore, the equation ofthe circle is
Question - 4 : - Find the equation of the circle with centre (1, 1) and radius 
Answer - 4 : -
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) =.
Therefore, the equation ofthe circle is
Question - 5 : - Find the equation of the circle with centre (–a, –b) and radius 
Answer - 5 : -
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) =.
Therefore, the equation ofthe circle is
Question - 6 : - Find the centre and radius of the circle (x +5)2 + (y – 3)2 = 36
Answer - 6 : -
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), while its radius is 6.
Question - 7 : - Find the centre and radius of thecircle x2 + y2 – 4x – 8y –45 = 0
Answer - 7 : -
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 = 
, which is of the form (x – h)2 + (y – k)2 = r2, where h =2, k = 4, and 
.
Thus,the centre of the given circle is (2, 4), while its radius is
.
Question - 8 : - Find the centre and radius of thecircle x2 + y2 – 8x +10y – 12 = 0
Answer - 8 : -
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
, which is of the form (x – h)2 + (y – k)2 = r2, where h =4, k = –5, and 
.
Thus,the centre of the given circle is (4, –5), while its radius is 
.
Question - 9 : - Find the centre and radius of the circle 2x2 + 2y2 – x =0
Answer - 9 : -
The equation of the given circle is 2x2 + 2y2 – x =0.
, which is of the form (x – h)2 + (y – k)2 = r2, where h = 
, k = 0, and 
.
Thus,the centre of the given circle is
, while its radius is 
.
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 - 10 : -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough 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 circlelies 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), weobtain h = 3 and k = 4.
On substituting the values of h and k inequation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough 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 circlelies 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), weobtain h = 3 and k = 4.
On substituting the values of h and k inequation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
Thus, the equation of therequired circle is
(x – 3)2 + (y –4)2 = 
x2 – 6x + 9+ y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y +15 = 0