Expert: Abe Mantell - 12/25/2009

Question
please can you explain to me how to solve these questions.thanks. 1, find the equation to the diameter of the circle x^2+y^2-8x+6y+21=0 which when produced passes through the point(2,1).                                                            2,  find the equation of the circle which passes through the point(1,1) has a radius 10, and whose centre lies on the line y=3x-7.

Answer
1. Do you mean the equation of the line that passes through (2,1) and the center of the circle?
If so, then "complete the square" to determine the center as follows:
x^2-8x+16-16+y^2+6y+9-9+21=0
(x-4)^2 + (y+3)^2 = 16+9-21
(x-4)^2+ (y+3)^2 = 4
So the center is at (4,3)

Now just find the equation of the line through (4,3) and (2,1)...OK?
I'll leave that for you to so.

2. We can find the center by finding the point(s) on the line y=3x-7 that is 10 units away from (1,1)
==> (x-1)^2 + (3x-7-1)^2 = 10^2 (via the distance formula)
==> 2x^2-10x-7=0
==> x=(5 + or - sqrt(39))/2 -- two possible centers on the line!

So there are two possible equations...now just use the formula:
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center & r=radius