c 1 : x 2 + y 2 − 2 x − 6 y + α = 0 {c_1}:{x^2} + {y^2} - 2x - 6y + \alpha = 0\\ c 1 : x 2 + y 2 − 2 x − 6 y + α = 0
Then centre = ( 1 , 3 ) = (1,3) = ( 1 , 3 ) ( r ) = 10 − α (r) = \sqrt {10 - \alpha } \\ ( r ) = 10 − α
Image of ( 1 , 3 ) (1,3) ( 1 , 3 ) x − y + 1 = 0 x - y + 1 = 0 x − y + 1 = 0 ( 2 , 2 ) (2,2)\\ ( 2 , 2 )
c 2 : 5 x 2 + 5 y 2 + 10 g x + 10 f y + 38 = 0 {c_2}:5{x^2} + 5{y^2} + 10gx + 10fy + 38 = 0\\ c 2 : 5 x 2 + 5 y 2 + 10 gx + 10 f y + 38 = 0
or x 2 + y 2 + 2 g x + 2 f y + 38 5 = 0 {x^2} + {y^2} + 2gx + 2fy + {{38} \over 5} = 0\\ x 2 + y 2 + 2 gx + 2 f y + 5 38 = 0
Then ( − g , − f ) = ( 2 , 2 ) ( - g, - f) = (2,2)\\ ( − g , − f ) = ( 2 , 2 )
∴ \therefore ∴ g = f = − 2 g = f = - 2 g = f = − 2
\\ c 2 = r = 4 + 4 − 38 5 = 10 − α {c_2} = r = \sqrt {4 + 4 - {{38} \over 5}} = \sqrt {10 - \alpha } \\ c 2 = r = 4 + 4 − 5 38 = 10 − α
⇒ 2 5 = 10 − α \Rightarrow {2 \over 5} = 10 - \alpha \\ ⇒ 5 2 = 10 − α
∴ \therefore ∴ α = 48 5 \alpha = {{48} \over 5} α = 5 48 r = 2 5 r = \sqrt {{2 \over 5}} \\ r = 5 2
∴ \therefore ∴ α + 6 r 2 = 48 5 + 12 5 = 12 \alpha + 6{r^2} = {{48} \over 5} + {{12} \over 5} = 12 α + 6 r 2 = 5 48 + 5 12 = 12
