How do you write the equation of the line that is tangent to the circle #(x + 6)^2 + (y + 4)^2 = 25# at the point (-9, -8)?

1 Answer
Jul 12, 2016

We will have to differentiate the function through implicit differentiation to find this one.

Let's first distribute:

#x^2 + 12x + 36 + y^2 + 8y + 16 = 25#

#x^2 + y^2 + 12x + 8y + 52 - 25 = 0#

#x^2 + y^2 + 12x + 8y + 27 = 0#

#x^2 + y^2 + 12x + 8y = -27#

We start by differentiating both sides:

#d/dx(x^2 + y^2 + 12x + 8y) = d/dx(-27)#

#2x + 2y(dy/dx) + 12 + 8(dy/dx) = 0#

#2y(dy/dx) + 8(dy/dx) = -12 - 2x#

#dy/dx(2y + 8) = -12 - 2x#

#dy/dx = (-12 - 2x)/(2y + 8)#

#dy/dx = -(2(6 + x))/(2(y + 4))#

#dy/dx = -(x + 6)/(y + 4)#

The slope of the tangent is given by inserting our given point, #(x, y)#, into the derivative.

Hence, #m_"tangent" = -(-9 + 6)/(-8 + 4) = -(-3/-4) = - 3/4#

The equation of the tangent can now be found by using point-slope form, since this will be a straight line.

#y - y_1 = m(x - x_1)#

#y - (-8) = -3/4(x - (-9))#

#y + 8 = -3/4x - 27/4#

#y = -3/4x - 27/4 - 8#

#y = -3/4x - 59/4#

So, the equation of the tangent to #y = (x + 6)^2 + (y + 4)^2 = 25# at the point #(-9, -8)# is #y = -3/4x - 59/4#.

Hopefully this helps!