Question

guys help me!
find the equation of the tangent line at (2,-1) on the circle (x+2)^2+(y+2)^2=25

Answer 1

we know that the centre of the circle is at (-2,-2) from equation of circle.
So the equation of the line (radius) joining (-2,-2) & (2,-1)
y+2/x+2=-1/-4

Answer 2

so y=x/4-3/2

Answer 3

so slope of radius=1/4
the slope of the tangent which is perpendicular to the radius will be -4 is y=-4x+c. now since it passes through (2,-1)----> -1=-4*2+c

Answer 4

(x+2)^2+(y+2)^2=25
take derivative
2(x+2)+2(y+2)dy/dx =0
dy/dx=-(x+2)/(y+2)
Slope at (2,-1)
m= dy/dx=-4
y=mx+c
y=-4x+c
-1=-4(2)+c
c=7
y=-4x+7
4x+y=7

Source: http://www.mathskey.com/question2answer/