Question

for the circle x^2+y^2+6x-4y+3=0 find the center and radius

Answer 1

x^2+y^2+6x-4y+3=0
x^2 +6x +9-9 + y^2 -4y -2 +2 +3=0
(x+3)^2 + (y-2)^2 = 9-2-3=4
so center at (-3,2) and radius =2

Answer 2

the equation of the tangent at (-2,5)

Answer 3

how to find this

Answer 4

sry, made a mistake:
x^2+y^2+6x-4y+3=0
x^2 +6x +9-9 + y^2 -4y +4 -4 +3=0
(x+3)^2 + (y-2)^2 = 9+4-3=10
so center at (-3,2) and radius =sqrt10

Answer 5

\[x^{2}+y^{2}+6x-4y+3\] -->
so just move the x's and y's together and put the constant on the other side

\[x^{2}+6x+y^{2}-4y=-3\]

[ (x^{2}+6x+9)+(y^{2}-4y+4)=-3+4+9\]

Answer 6

i got u @Fgcbear16 and @myko but now i stuck with the tangent

Answer 7

the equation of the tangent at (-2,5)

Answer 8

good luck!

Answer 9

x^2+y^2+6x-4y+3=0
derivate this implicitly:
2x+2yy' +6-4y'=0
put terms with y' one side and the rest other:
y'(2y-4)=-2x-6
y'=(-2x-6)/(2y-4)
substitute value of the point (-2,5)
y'=(4-6)/(10-4)=-2/6=-1/3
this is your slope. I think you can finish alone,:)