NCERT Solutions
Class 11 Maths
Conic Sections
Ex.11.1 Q.13
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
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 = 
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 = 
Thus, the equation of the required circle is
(x – )2 + (y – )2 = ( )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