How do you find the equations of common tangents to the circles #x^2+y^2=9, x^2+y^2-16x+2y+49=0#?

1 Answer
Shwetank Mauria
Jun 30, 2017

Tangents are #y=3#; #63y+16x+195=0#; #12x+5y-39=0# and #4x-3y-15=0#

Explanation:

First circle is #x^2+y^2=9#, whose center is #(0,0)# and radius is #3# and second circle is #x^2+y^2-16x+2y+49=0#, whose center is #(8,-1)# and radius is #sqrt(8^2+(-1)^2-49)=sqrt16=4#.

As distance between centers is #sqrt(8^2+1^2)=sqrt65# and is greater than the sum of the radius, which is #7#, the circles do not overlap. As such, we will have four tangents, two transverse and two direct.

The point of intersection of transverse common tangents will intersect each other at a point on the line joining centers and will internally divide the line in the ratio of their radii. And point of intersection of direct common tangents will intersect each other at a point on the line joining centers and will externally divide the line in the ratio of their radii.

Hence point of intersecrtion of transverse tangents is #((0xx4+8xx3)/7,(0xx4+(-1)xx3)/7)# i.e. #(24/7,-3/7)# and tangents are #(y+3/7)=m(x-24/7)#

Similarly point of intersecrtion of direct tangents will be #((0xx4-3xx8)/(4-3),(0xx4-3xx(-1))/(4-3))# i.e. #(-24,3)# and tangents are #(y-3)=m(x+24)#

To find the slopes, although it is a bit longer, but I refer to this page , which gives the slope of tangent from an external point #(p,q)# to a circle #x^2+y^2=R^2# as #(-pq+-Rsqrt(p^2+q^2-R^2))/(R^2-p^2)# and using this we can find values of #m#'s using first external point #(-24,3)# and then internal point as #(24/7,-3/7)#. In both cases #R=3#.

Hence value of #m# for #(-24,3)# is

#(-(-72)+-3xxsqrt(576+9-9))/(9-576)# or #(72+-72)/(-567)# or #0# or #-16/63#

and tangents using #(y-3)=m(x+24)# are

#y=3# and #63y+16x+195=0#

Similary for #(24/7,-3/7)# slopes are given by #(72/49+-3xxsqrt(576/49+9/49-9))/(9-576/49)# or #(72+-3/7sqrt(576+9-441))/((441-567)/49)#,
which on simplification gives slopes as #-12/5# and #4/3#

and tangents using #(y+3/7)=m(x-24/7)# are

#12x+5y-39=0# and #4x-3y-15=0#

graph{(x^2+y^2-9)(x^2+y^2-16x+2y+49)(4x-3y-15)(12x+5y-39)(63y+16x+195)(y-3)=0 [-6.08, 13.92, -5.32, 4.68]}

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