Find the equations of the external common tangents and internal common tangents to these circles?

:#x^2 + y^2 + 10x + 21 = 0# and #x^2 + y^2 - 10x - 39 = 0#

1 Answer
Mar 12, 2018

Internal tangents is #x+3=0# and external tangents are #4y=3x+25# and #4y=-3x-25#

Explanation:

As center of circle #x^2+y^2+2gx+2fy+c=0# is #(-g,-f)# andradius is #sqrt(g^2+f^2-c)#

Hence center of circle #x^2+y^2+10x+21=0# is #(-5,0)# and radius is #sqrt((-5)^2+0^2-21)=sqrt4=2#

and for #x^2+y^2-10x-39=0# center is #(5,0)# and radius is #sqrt(5^2+0^2-(-39))=sqrt64=8#.

It is observed that as distance between centers is #10# (both lie on #x#-axis so it is easy) and this is equal to sum of radii, the two circles touch each other on #x#-axis at point #(-3,0)# and therefore

there would be only one common tangent, whose equation is #x+3=0#.

And point of intersection of external common tangents will be on the line joining centers #(-5,0)# and #(5,0)# and will externally divide the line in the ratio of their radii i.e. in the ratio #2:8#. This will be

#((2xx5-8xx(-5))/(2-8),(2xx0-8xx0)/(2-8))# i.e. #(-50/6,0)# or #(-25/3,0)# i.e. its equation (using point slope form of equation) would be #y=m(x+25/3)#, where #m# is the slope of tangent.

This tangent touches any circle say #x^2+y^2+10x+21=0# at one point. Therefore putting #y=m(x+25/3)# in this equation we get

#x^2+m^2(x+25/3)^2+10x+21=0#

or #9x^2+9m^2x^2+150m^2x+625m^2+90x+189=0#

or #x^2(9+9m^2)+x(150m^2+90)+(625m^2+189)=0#

As there is only one point the discriminant must be zero i.e.

#(150m^2+90)^2-4(9+9m^2)(625m^2+189)=0#

or #22500m^4+27000m^2+8100-22500m^4-29304m^2-6804=0#

or #-2304m^2+1296=0#

or #16m^2-9=0#

i.e. #m=+-3/4# and equation of tangents are

#y=3/4(x+25/3)# or #4y=3x+25#

and #y=-3/4(x+25/3)# or #4y=-3x-25#

graph{(x^2+y^2+10x+21)(x^2+y^2-10x-39)(x+3)(4y-3x-25)(4y+3x+25)= 0 [-20, 20, -10, 10]}