SOLUTION: Find the equation of the tangent to the circle with centre: (0,2) at (-1,5)

Question 770881: Find the equation of the tangent to the circle with centre: (0,2) at (-1,5)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the equation of the tangent to the circle with centre: (0,2) at (-1,5)
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radius = sqrt[(5-2)^2+(-1-0)^2] = sqrt[10]
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Equation of the circle:
x^2 + (y-2)^2 = 10
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Solve for "y":
y^2-4y+4 + x^2 = 10
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Find dy/dx
2y(dydx) -4 + 2x = 0
dydx = (-2x+4)/(2y)
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dy/dx at (-1,5) is :: (2+4)/-2 = -3
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Equation of the line with slope = -3 passing thru (-1,5)
y = mx + b
Solve for "b":
5 = -3*-1 + b
b = 2
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Equation:
y = -3x + 2
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Cheers,
Stan H.
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