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+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).

- 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 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

- 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 boo-boos help keep us humble...

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