A-maths circle

Firstly, the locus of P is actually a circle and its centre is ( 3 , 2 ).
Rearranging,
x2 + y2 - 6x - 4y + 14 = k
x2 + y2 – 6x – 4y + ( 14 – k ) = 0
If the circle touches the 4 lines, then the distance between the line and the centre of the circle is the radius.
r = ( 1 / 2 )√[ ( - 6 )2 + ( - 4 )2 – 4 ( 14 – k )]
r = √( k – 1 )
Consider L4,
d = l ( Ax1 + By1 + C ) / √( A2 + B2 ) l
√( k – 1 ) = l [ ( 1 )( 3 ) + ( - 1 )( 2 ) ] / √[ ( - 1 )2 + ( - 1 )2] l
√( k – 1 ) = l 1 / √2 l
k – 1 = 1 / 2
k = 1.5


2007-09-30 18:19:50 補充：
Sorry for a typing error, it should be consider L1, but not L4.

2007-09-30 18:24:13 補充：
Sorry again, the above is correct, i.e. consider L4. Actually no matter which lines you are considering, you can reach the same answer. Say L1: √( k – 1 ) = l [( 3 )( 1 ) + ( 2 )( 1 ) - 4] / √( 1^2 + 1^2 )l k - 1 = ( 1 / √2 )^2 k - 1 = 1 / 2 k = 1.5

2007-09-30 18:25:51 補充：
For L2: √( k – 1 ) = l [ ( 3 )( 1 ) + ( 2 )( - 1 ) - 2 ] / √[ 1^2 + ( - 1 )^2 ]lk - 1 = ( - 1 / √2 )^2 k - 1 = 1 / 2 k = 1.5

2007-09-30 18:30:13 補充：
For L3: √( k – 1 ) = l [ ( 3 )( 1 ) + ( 2 )( 1 ) - 6 ] / √( 1^2 + 1^2 )l k - 1 = ( - 1 / √2 )^2 k - 1 = 1 / 2k = 1.5But surely considering L4 is the simpliest way to reach the answer.

The centre of the circle P is (3, 2) and the radius = sqrt(3^2+2^2-(14-k)) = sqrt(k-1)

Distance between the centre and L1 = |(3+2-4)/sqrt(1+1)| = 1/sqrt2
Similarly, the distances between the centre and the other three lines are also 1/sqrt2
Hence if the circle touches the four lines, then
sqrt(k-1) = 1/sqrt2
2(k-1) = 1
k = 3/2