The circle x 2 + y 2 + 6 x + 8 y + 16 = 0 {x^2} + {y^2} + 6x + 8y + 16 = 0 x 2 + y 2 + 6 x + 8 y + 16 = 0 ( − 3 , − 4 ) ( - 3, - 4) ( − 3 , − 4 )
The circle
x 2 + y 2 + 2 ( 3 − 3 ) x + 2 ( 4 − 6 ) y = k + 6 3 + 8 6 , k > 0 {x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y \\ = k + 6\sqrt 3 + 8\sqrt 6 ,\,k > 0 x 2 + y 2 + 2 ( 3 − 3 ) x + 2 ( 4 − 6 ) y = k + 6 3 + 8 6 , k > 0 has centre
( 3 − 3 , 6 − 4 ) \left( {\sqrt 3 - 3,\,\sqrt 6 - 4} \right) ( 3 − 3 , 6 − 4 ) and radius
k + 34 \sqrt {k + 34} k + 34 ∵ \because ∵
3 + 6 = ∣ k + 34 − 3 ∣ \sqrt {3 + 6} = \left| {\sqrt {k + 34} - 3} \right| 3 + 6 = k + 34 − 3 Here, k = 2 k = 2 k = 2 ∵ k > 0 \because k > 0 ∵ k > 0
Equation of common tangent to two circles is
2 3 x + 2 6 y + 16 + 6 3 + 8 6 + k = 0 2\sqrt 3 x + 2\sqrt 6 y + 16 + 6\sqrt 3 + 8\sqrt 6 + k = 0 2 3 x + 2 6 y + 16 + 6 3 + 8 6 + k = 0
then equation is
x + 2 y + 3 + 4 2 + 3 3 = 0 . . . . . . ( i ) x + \sqrt 2 y + 3 + 4\sqrt 2 + 3\sqrt 3 = 0 \quad ...... (i) x + 2 y + 3 + 4 2 + 3 3 = 0 ...... ( i ) ∵ ( α , β ) \because (\alpha, \beta) ∵ ( α , β ) ( − 3 , − 4 ) ( - 3, - 4) ( − 3 , − 4 )
To line (i) then
α + 3 1 = β + 4 2 = − ( − 3 − 4 2 + 3 + 4 2 + 3 3 ) 1 + 2 {{\alpha + 3} \over 1} = {{\beta + 4} \over {\sqrt 2 }} \\ = {{ - \left( { - 3 - 4\sqrt 2 + 3 + 4\sqrt 2 + 3\sqrt 3 } \right)} \over {1 + 2}} 1 α + 3 = 2 β + 4 = 1 + 2 − ( − 3 − 4 2 + 3 + 4 2 + 3 3 )
∴ α + 3 = β + 4 2 = − 3 \therefore \alpha + 3 = {{\beta + 4} \over {\sqrt 2 }} = - \sqrt 3 ∴ α + 3 = 2 β + 4 = − 3
⇒ ( α + 3 ) 2 = 9 \Rightarrow {\left( {\alpha + \sqrt 3 } \right)^2} = 9 ⇒ ( α + 3 ) 2 = 9 and
( β + 6 ) 2 = 16 {\left( {\beta + \sqrt 6 } \right)^2} = 16 ( β + 6 ) 2 = 16
∴ ( α + 3 ) 2 + ( β + 6 ) 2 = 25 \therefore {\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2} = 25 ∴ ( α + 3 ) 2 + ( β + 6 ) 2 = 25 
