Posted 17 April 2004 - 02:23 PM

For 2b, the question says the line QC is parallel to the y-axis. That means that since the point Q has x-cordinate 1, the point and centre C will also have the x-cordinate of 1.

Since C lies on the line AB, you can substitute that into the forumla from part a, x + 2y = 9 so 1 + 2y = 9 leading to y = 4

So the centre of the circle is (1,4)

Then to find the radius of the circle you can use the distance formula to find the length of QC, which is the radius.

CQ = √(1-1)² + (9-4)²

CQ = √(0)² + (5)²

CQ = √25

CQ = 5

Put that into the equation of the circle:

( x - a )² + ( y - b )² = r²

( x - 1 )² + ( y - 4 )² = 5²

( x - 1 )² + ( y - 4 )² = 25

You could multiply that out but you don't have to