What is the equation of the tangent at (0 , 2) to the circle with equation (x + 2)^2 + (y + 1)^2 = 13?

Answer 1

The center of the circle is at (-2,-1). We know that the radius of the circle goes through (0,2). You can derive the straight line equation of this radius line. But all you need is the slope of this radius line which is

m_r = (2+1)/(0+2) = 3/2. The tangent line is perpendicular to this radius line. So, the slope of the tangent

line is -1/(3/2) = -2/3. Therefore, the equation of the tangent line going through

point (0,2)

y-2 = (-2/3)(x-0) or y = (-2/3)x + 2