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
or
-x + y = -1
This intersects the line
x + y = 3 at
(2,1)
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.
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).
1 answer
2 answers