Question

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Answer 1

The given point ( 0,0 ) lie on the circle which makes intercepts a and b on the coordinate axes

The equation of the required circle is given by,

( x−h ) 2 + ( y−k ) 2 = r 2 (1)

Where h and k denotes the center of the circle and r denotes the radius of the circle.

Since circle also passes through the point ( 0,0 ) , this point will satisfy the equation of the circle represented in equation ( 1 )

( 0−h ) 2 + ( 0−k ) 2 = r 2 h 2 + k 2 = r 2

The equation of the circle becomes,

( x−h ) 2 + ( y−k ) 2 = h 2 + k 2 (2)

It is given that the circle makes a intercept with coordinate axes. This means that the circle passes through point (a,0) . So,

( a−h ) 2 + (0−k) 2 = h 2 + k 2 a 2 −2ah+ h 2 + k 2 = h 2 + k 2 a 2 −2ah=0 a( a−2h )=0

Further simplify equations,

a=0 or a−2h=0 h= a 2

Since a≠0 , implies h= a 2

Similarly, it is given that the circle makes b intercept with coordinate axes. This means that the circle passes through point (0,b) . So,

( 0−h ) 2 + (b−k) 2 = h 2 + k 2 h 2 + b 2 −2bk+ k 2 = h 2 + k 2 b 2 −2bk=0 b( b−2k )=0

Further simplify the equations,

b=0 or k= b 2

Since b≠0 implies k= b 2

Substitute the values oh h and k in equation (2).

( x− a 2 ) 2 + ( y− b 2 ) 2 = ( a 2 ) 2 + ( b 2 ) 2 ( 2x−a 2 ) 2 + ( 2y−b 2 ) 2 =( a 2 + b 2 4 ) 4 x 2 −4ax+ a 2 +4 y 2 −4by+ b 2 = a 2 + b 2 x 2 −ax+ y 2 −by=0

Thus the equation of the circle passing through ( 0,0 ) and making intercepts a and b on the coordinate axes is x 2 −ax+ y 2 −by=0 .