math
posted by steev on .
can u help
this the qustion is :
A circle passes through the points (1,4)&(5,0) and its centre lies on the line x+y3=0.Find
(i) the equation of the circle and it parametric equations .
(ii) the centre,diameter and area of the circle .
(iii) the equation of the tangent at (1,4) .
(iv) the equation of the circle concentric with the above circle and passing through the point (7,8).

let A be (1,4) and B be (5,0)
So AB must be a chord of the circle and the centre must be the intersection of the rightbisector of AB and the line x+y  3 = 0
slope of AB = 4/4 = 1
slope of rightbisector = 1
midpoint of AB = (3,2)
equation of rightbisector: y = x + b
with (3,2) on it
2 = 3 + b > b = 1
rightbisector: y = x1
solve with the second equation:
x+y  3 = 0
x + (x1) = 3
2x = 4
x = 2, then y = 3
centre is (2,3)
radius is distance from centre to A
= √((21)^2 + (34)^2) = √2
i) equation of circle: (x2)^2 + (y3)^2 = 2
ii) you do it
iii) slope of radius from A to centre = ...
slope of tangent at A is ..... (the negative reciprocal of above
now you have a point and slope of the tangent....
iv) equation will be
(x2)^2 + (y3)^ = r^2
plug in (7,8) to get r^2 
Just noticed that you posted the same question twice, and that both Steve and I answered it.
BUT, I made a stupid error in my centre calculation, (that's what you get when you try to do things in your head without writing them down)
I had the centre as (2,3), should have been (2,1) like Steve had
so go with his solution. 
Heh heh  We all do these problems just for fun anyway. Any help to the students is just a welcome side effect. And, these little booboos help keep us humble...