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For the circle x^2+y^2+6x-4y+3=0
how would i go about finding:
a) the center and the radius
b) the equation of the tangent line at the point (-2,5)

any help is mch appreciated. thanks!

  • math -

    (a) Rewrite the equation as

    x^2 + 6x + 9 + y^2 -4y + 4 +3 -13 = 0
    (x+3)^2 + (y-2)^2 = 10 = (sqrt 10)^2

    The should tell you that the center is at (-3, 2) and the circle radius is sqrt 10.

    (b) First cmpute the slope dy/dx at (-2, 5)using implicit differentiation.

    2x + 2y y' + 6 -4 y'= 0
    y'(4 - 2y) = 2x + 6
    y' = 2/(4-2y) = -1/3

    Finally, write down the equation of s straight line passing through (-2,5) with that slope.

    Verify my numbers. I may have goofed somewhere along the line

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