Slope of tangent to the circle #(x-r)^2 + y^2 = r^2# at the point #(x,y)# lying on the circle is?

Question

A) #x/(y-r)#
B) #(r-x)/y#
C) #(y^2-x^2)/y#
D) #(y^2 + x^2)/(2xy)#

2274 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-20T17:47:12

B) #(r-x)/y#

Implicitly differentiate #(x-r)^2 + y^2 = r^2#

#(d((x-r)^2))/dx + (d(y^2))/dx = (d(r^2))/dx#

#2(x-r) + 2ydy/dx = 0#

#x-r + ydy/dx = 0#

#ydy/dx = r-x#

#dy/dx = (r-x)/y#

This is selection B)