Asked by Jack
                Find the equation of the diameter of the circle x^2+y^2-6x+2y=15, which, when produced, passes through the point (8,-2).
            
            
        Answers
                    Answered by
            Reiny
            
    completing the square will give us the centre ....
x^2 - 6x + .... + y^2 + 2y + ... = 15
x^2 - 6x + 9 + y^2 + 2y + 1 = 15 + 9+1
(x-3)^2 + (y+1)^2 = 25
centre is (3,-1)
so you want the line through (3,-1) and (8,-2)
slope = (-2+1)/(8-3) = -1/5
y+1 = (-1/5(x-3)
5y + 5 = -x + 3
x + 5y = -2 or y = (-1/5)x - 2/5
    
x^2 - 6x + .... + y^2 + 2y + ... = 15
x^2 - 6x + 9 + y^2 + 2y + 1 = 15 + 9+1
(x-3)^2 + (y+1)^2 = 25
centre is (3,-1)
so you want the line through (3,-1) and (8,-2)
slope = (-2+1)/(8-3) = -1/5
y+1 = (-1/5(x-3)
5y + 5 = -x + 3
x + 5y = -2 or y = (-1/5)x - 2/5
                    Answered by
            Jack
            
    thanks
    
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