posted by math 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).
The center of the circle also lies on the perpendicular bisector of any chord. We have the chord between the two given points.
The chord has slope (0-4)/(5-1) = -1.
The midpoint of the chord is (3,2).
The equation of the line containing (3,2) and perpendicular to the chord is
(y-2)/(x-3) = 1
-x + y = -1
This intersects the line
x + y = 3 at
So, our circle has equation
(x-2)^2 + (y-1)^2 = r^2
plug in either of the given points and we find r^2 = 10. so,
(x-2)^2 + (y-1)^2 = 10
x = 2 + √10 cos t
y = 1 + √10 sin t
the rest should be easy to figure out.