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+y-3=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 right-bisector of AB and the line x+y - 3 = 0
slope of AB = -4/4 = -1
slope of right-bisector = 1
midpoint of AB = (3,2)
equation of right-bisector: y = x + b
with (3,2) on it
2 = 3 + b ---> b = -1
right-bisector: y = x-1
solve with the second equation:
x+y - 3 = 0
x + (x-1) = 3
2x = 4
x = 2, then y = 3
centre is (2,3)
radius is distance from centre to A
= √((2-1)^2 + (3-4)^2) = √2
i) equation of circle: (x-2)^2 + (y-3)^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
(x-2)^2 + (y-3)^ = 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 boo-boos help keep us humble...