How do you write an equation for the line tangent to the circle x^2 + y^2 = 29 at the point (2, 5)?

1 Answer
May 11, 2016

Answer:

#2x+5y=29#

Explanation:

If #x^2+y^2=29#
then by taking the derivative of both sides (to obtain the general slope)
#color(white)("XXX")2x+2y*(dy)/(dx)=0#
and
the slope is
#color(white)("XXX")m=(dy)/(dx)=-x/y#

Specifically the slope at #(x,y)=(2,5)#
will be #m=-2/5#
and using the slope point form for the tangent:
#color(white)("XXX")(y-5)=-2/5(x-2)#

#color(white)("XXX")5y-25=-2x+4#

#color(white)("XXX")2x+5y=29# (after conversion to standard form)

Here is a graph of the (top half of the) circle equation and the calculated tangent line.
graph{(x^2+y^2-29)(2x+5y-29)=0 [-7.58, 8.224, -0.3, 7.6]}

(note: I am having some problems with this graph display).