posted by alk on .
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!
(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