equation of circle

( a )

F: x² + y² + ( 2k - 6 )x + ( 4k - 4 )y + ( k - 2 ) = 0

centre
= ( -( 2k - 6 ) / 2 , -( 4k - 4 ) / 2 )
= ( 3 - k, 2 - 2k )

radius
= ( 1 / 2 )√[( 2k - 6 )² + ( 4k - 4 )² - 4( k - 2 )]
= ( 1 / 2 )√[( 4k² - 24k + 36 ) + ( 16k² - 32k + 16 ) - 4( k - 2 )]
= ( 1 / 2 )√( 4k² - 24k + 36 + 16k² - 32k + 16 - 4k + 8 )
= ( 1 / 2 )√( 20k² - 60k + 60 )
= √( 5k² - 15k + 15 )


( b )

5k² - 15k + 15
= 5( k² - 3k ) + 15
= 5[ k² - 3k + ( 3 / 2 )²] - 5( 3 / 2 )² + 15
= 5( k - 3 / 2 )² + 15 / 4
thus, when k = 3 / 2, the radius attains the minimum.

the minimum radius
= √[ 5( 3 / 2 )² - 15( 3 / 2 ) + 15 ]
= √( 15 / 4 )
= √15 / 2

AB = 2 x the minimum radius = √15


( ci )

the distance from the centre to L
= | 4( 3 - k ) - 2( 2 - 2k ) + 7 | / √( 4² + 2² )
= | 12 - 4k - 4 + 4k + 7 | / √20
= 15 / √20
= 3√5 / 2 ( constant )

the locus of the centres of F is parallel to L.


( cii )

Let r be the required radius
According to Pyth. theorem,
r² = ( 3√5 / 2 )² + ( √15 / 2 )²
r² = 45 / 4 + 15 / 4
r² = 15
r = √15

set √( 5k² - 15k + 15 ) = √15
5k² - 15k + 15 = 15
5k² - 15k = 0
k( k - 3 ) = 0
k = 0 or 3

thus, the two possible circles are
x² + y² + ( 2( 0 ) - 6 )x + ( 4( 0 ) - 4 )y + (( 0 ) - 2 ) = 0 and
x² + y² + ( 2( 3 ) - 6 )x + ( 4( 3 ) - 4 )y + (( 3 ) - 2 ) = 0
i.e. x² + y² - 6x - 4y - 2 = 0 and x² + y² + 8y + 1 = 0

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