Question

find the equation of the circle which passes through the points ( 5 , 0 ) and ( 1 , 4 ) and whose centre lies on the line x + y - 3 = 0 .

Answer 1

Heya User,

--> Let the center be O( x , y )

--> By definition of Circle -->
 ----------> Distance b/w ( 5 , 0 ) and O = Distance b/w ( 1 , 4 ) and O

=> [ x - 5 ]² + y² = [ x - 1 ]² + [ y - 4 ]²
=> -10x + 25 = -2x + 1 - 8y + 16

=> 8x - 8y = 8
=> x - y = 8

However, Center lies on the line --> x + y - 3 = 0 

=> O must lie on the intersection of x - y = 8 and x + y = 3

--> ( x , y ) ≡ ( 11/2 , -5/2 )

--> r² --> [ ( 1/2 )² + ( 5/2 )² ] = [ 26 / 4 ]

Hence, the eqn. -->

---> [ x - 11/2 ]² + [ y + 5/2 ]² = [ 26 / 4 ]
--> [ 2x - 11 ]² + [ 2y + 5 ]² = 26    ^_^ Done..