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#1
Finding tangent to circle x^2+y^2+2gx+2fy+c=0, at point (h,k)

The book says hx+ky+g(x+h) + f(y+k) + c =0

but i keep getting hx+ky+g(x-h) + f(y-k) + c =0

Where am I going wrong?
0
14 years ago
#2
2x + 2y dy/dx + 2g + 2f dy/dx = 0
dy/dx = -(x + g)/(y + f).

So at (h, k) the gradient is -(h + g)/(k + f) and the tangent is

y = -(h + g)x/(k + f) + k + (h + g)h/(k + f)
(k + f)y = -(h + g)x + k (k + f) + (h + g)h
(h + g)x + (k + f)y = k (k + f) + (h + g)h
(h + g)x + (k + f)y = (h^2 + k^2 + 2gh + 2fk) - fk - gh
(h + g)x + (k + f)y = -c - fk - gh . . . using the circle equation
(h + g)x + (k + f)y + fk + gh + c = 0
hx + ky + g(x + h) + f(y + k) + c = 0.
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