Posted by steev on Wednesday, December 28, 2011 at 12:34am.
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).

math  Reiny, Wednesday, December 28, 2011 at 8:44am
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

ARghh math  Reiny, Wednesday, December 28, 2011 at 9:19am
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.

math  to Reiny  Steve, Wednesday, December 28, 2011 at 11:03am
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...
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