Chapter 11 Conic Sections Ex 11.1 Solutions
Question - 11 : - Find the equation of the circle passingthrough the points (2, 3) and (–1, 1) and whose centre is on the line x –3y – 11 = 0.
Answer - 11 : -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough 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 circlelies on line x – 3y – 11 = 0,
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 
.
On substituting the values of h and k inequation (1), we obtain
Thus, the equation of therequired circle is
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 - 12 : -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the radius of the circle is 5 and itscentre 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 thecircle passes through point (2, 3).
When h = –2, the equationof the circle becomes
(x + 2)2 + y2 = 25
x2 + 4x + 4+ y2 = 25
x2 + y2 + 4x –21 = 0
When h = 6, the equation ofthe 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 - 13 : -
Let the equation of the requiredcircle be (x – h)2 + (y – k)2 = r2.
Sincethe circle passes through (0, 0),
(0– h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
Theequation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
Itis given that the circle makes intercepts a and b onthe 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)
Fromequation (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 =
.
Fromequation (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 = 
.
Thus,the equation of the required circle is
Question - 14 : - Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Answer - 14 : -
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).
Thus, the equation of the circle is
Question - 15 : - Does the point (–2.5, 3.5) lie inside,outside or on the circle x2 + y2 = 25?
Answer - 15 : -
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.
∴Centre= (0, 0) and radius = 5
Distance between point(–2.5, 3.5) and centre (0, 0)
Sincethe distance between point (–2.5, 3.5) and centre (0, 0) of the circle is lessthan the radius of the circle, point (–2.5, 3.5) lies inside the circle.