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December 21, 2014

December 21, 2014

Posted by **math** on Wednesday, December 28, 2011 at 12:33am.

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).

- algebra -
**Steve**, Wednesday, December 28, 2011 at 4:32amThe 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.

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