equation of circle

We may tackle the question as follows:

Step 1: Construct a useful quadratic equation
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Rewrite x + 3y - 11 = 0 as x = 11 - 3y first.

Substitute x = 11 - 3y into the circle equation to form a quadratic equation in y ( the two roots are the y-coordinates of the two intersection points between the chord and the circle itself )

[ ( 11 - 3y ) - a ]^2 + ( y - a )^2 = 5
[ ( 11 - a ) - 3y ]^2 + ( y - a )^2 = 5
( 11 - a )^2 - 6( 11 - a )y + 9y^2 + ( y^2 - 2ay + a^2 ) = 5
( 121 - 22a + a^2 ) - 6( 11 - 3a )y + 9y^2 + ( y^2 - 2ay + a^2 ) = 5
10y^2 + ( 4a - 66 )y + ( 116 - 22a + 2a^2 ) = 0 ---- (*)


Step 2: Use the coordinate of their intersection point to find "a"
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There is no need to solve it directly for it involves "a".
We can consider finding the necessary coordinate by using the sum and product of roots which we explore in S.4.:

As M( 3.5, 2.5 ) is the midpoint of two intersection points AB,
Sum of two y-coordinates of A and B = 2( y-coordinate of M )
-( 4a - 66 )/10 = 2( 2.5 )
66 - 4a = 10( 5 ) = 50
4a = 16
a = 4
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Such "a" can further tell you the coordinate of centres and you can also solve the two intersection points easily by using (*) above. All depend on whether the question has asked or not.

Hope I have helped you.

翻雷滾天 風卷殘雲 ,你太心急了.
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