Kaustubh Nayyar
Last Activity: 10 Years ago
hey I have a simpler and unique solution Use family of circles passing through two points and then put third point to get the constant
The required circle’s equation will be S + k L = 0
where S is equation of circle passing through any two points ( lets say (6,0) and ( – 2, – 4) ) as diametric points k is the constant we have to find and L = 0 is the equation of diameter.
hence eq. of reqd. circle will be
(x – 6)(x + 2 ) + (y – 0)(y + 4 ) + k (2y – x + 6 ) = 0
Now, ( 1, 5) lies on the reqd. circle
hence we get k = – 2
therefore, eq of reqd. circle (x – 6)(x + 2 ) + (y – 0)(y + 4 ) – 2 (2y – x + 6 ) = 0
which gives x2 + y2 – 2x – 24 = 0
( this is a unique solution , you cannot find this method in books )