Find the equation of the tangent at the point (0 2) to the circle (x+2)^2+(y+1)^2=13
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(x+2)^2+(y+1)^2=13
diff.both side wrt x
2(x+2)+2(y+1)dy/dx=0
dy/dx=-2(x+2)/2(y+1)
dy/dx=-(x+2)/y+1
at point (0,2)dy/dx=-2/3
equation of tangent=y-2=-2/3(x-0)
y-2=-2/3(x)
3y-6=-2x
3y+2x-6=0
diff.both side wrt x
2(x+2)+2(y+1)dy/dx=0
dy/dx=-2(x+2)/2(y+1)
dy/dx=-(x+2)/y+1
at point (0,2)dy/dx=-2/3
equation of tangent=y-2=-2/3(x-0)
y-2=-2/3(x)
3y-6=-2x
3y+2x-6=0
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