f.5 maths (locus)
2008-09-18 8:38 am
find the equation of the tangent to the circle x^2+y^2-2x+6y+5=0 at the point (3,-2)
回答 (2)
2008-09-18 8:50 am
✔ 最佳答案
Differentiate x^2+y^2-2x+6y+5=0 with respect to x,2x + 2y dy/dx - 2 + 6 dy/dx = 0
dy/dx = (1-x) / (3 + y)
Slope of tangent at (3,-2) = ( 1 - 3 ) / ( 3 - 2) = -2
So equation of tangent is
( y + 2 ) / (x - 3) = -2
y + 2 = 6 - 2x
2x + y - 4 = 0
2008-09-18 3:15 pm
For point P(x1, y1) on a circle, equation of tangent to the circle x^2 + y^2 + Dx + Ey + F = 0 at P is given by the formula:
x1x + y1y + D(x +x1)/2 + E(y1 + y)/2 + F = 0
Now x1 = 3 and y1 = -2.
Therefore equation of tangent is
3x - 2y -2(3 + x)/2 + 6(-2 + y)/2 + 5 = 0
3x - 2y - 3 - x - 6 + 3y + 5 = 0
2x + y = 4.
x1x + y1y + D(x +x1)/2 + E(y1 + y)/2 + F = 0
Now x1 = 3 and y1 = -2.
Therefore equation of tangent is
3x - 2y -2(3 + x)/2 + 6(-2 + y)/2 + 5 = 0
3x - 2y - 3 - x - 6 + 3y + 5 = 0
2x + y = 4.
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