Question

How do you find the equations of common tangents to the circles `x^2+y^2-2x-6y+6=0, x^2+y^2=1`?

Answer 1

Direct tangents are `x+1=0` and `4x-3y-5=0` and transverse tangents are `y-1=0` and `3x+4y-5=0`

Explanation:

The center of `x^2+y^2-2x-6y+6=0` is `(1,3)` and radius is `sqrt(1^2+3^2-6)=2` and center and radius of `x^2+y^2=1` is `(0.0)` and `1`. As distance between centers is `sqrt10=3.162` is greater than sum of the radii, 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 `((0xx2+1xx1)/3,(0xx2+3xx1)/3)` i.e. `(1/3,1)` and tangents are `(y-1)=m(x-1/3)` or `3y-3=m(3x-1)`

Similarly point of intersecrtion of direct tangents will be `((0xx2-1xx1)/(2-1`),(0xx2-1xx3)/(2-1))` i.e. `(-1,-3)` and tangents are `(y+3)=m(x+1)`

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 `(-1,-3)` and then internal point as `(1/3,1)`. In both cases `R=1`.

As for `(-1,-3)`, we have `R^2-p^2=0`, the value of `m` is

`oo` and `(-q^2+R^2)/(-2Rq)` i.e. `-(-9+1)/6` or `4/3`

and tangents are `x+1=0` and `y+3=4/3(x+1)` i.e. `4x-3y-5=0`

Similary for `(1/3,1)` slopes are given by `(-1/3+-sqrt(1/9+1-1))/(1-1/9)` or `(-1/3+-1/3)/(8/9)`, i.e. `0` and `-3/4`

and tangents are `y-1=0` and `y-1=-3/4(x-1/3)` or `3x+4y-5=0`

graph{(x^2+y^2-1)(x^2+y^2-2x-6y+6)(y-1)(3x+4y-5)(x+1)(4x-3y-5)=0 [-9.96, 10.04, -3.4, 6.6]}

Answer 2

`4x-3y-5=0.`

`y-1=0.`

`3x+4y-5=0.`

`x+1=0.`

Explanation:

Let the given Circles be,

`S_1 : x^2+y^2=1, and, S_2 : x^2+y^2-2x-6y+6=0.`

The `" Centres "C_i," and, Radii "r_i" of "S_i, i=1,2,` are,

`C_1(0,0), C_2(1,3), r_1=1, and, r_2=2.`

Let, `y=mx+c` be the eqn. of common tgt. of `S_i, i=1,2.` Then,

from Geometry, we know that, the `bot"-dist."` from centre to tgt.

line is same as radius.

Hence, for `S_1," we have, "|c|/sqrt(1+m^2)=1, or, c^2=1+m^2.......(1).`

For `S_2, |m-3+c|/sqrt(1+m^2)=2, or, {m+(c-3)}^2=4(1+m^2)...(2).`

`(1) & (2) rArr {m+(c-3)}^2=4c^2 rArr m+c-3=+-2c.`

`m+c-3=2c rArr m=3+c.....................................(3),` and,

`m+c-3=-2c rArr m=3-3c.............................(4).`

Then, `(1), &, (3) rArr c^2=1+9+6c+c^2.`

`:.c=-5/3, m=3-5/3=4/3............(5).`

Similarly, `(1),&,(4) rArr c^2=1+(3-3c)^2=10-18c+9c^2.`

`:. 8c^2-18c+10=0 rArr (c-1)(8c-10)=0.`

`:. c=1, m=3-3c=0, &, c=5/4, m=-3/4...............(6).`

`(5) and (6)` give us the following pairs,

`(m,c)=(4/3,-5/3), (0,1) and, (-3/4,5/4).`

Only thing that remains is to check whether there exists any

Vertical common tgt. with eqn. like, `x=k.........(7).`

So, we do the same exercise with `(7)` as we have done above

with `y=mx+c.`

`"For "S_1, |k|=1rArrk=+-1,"` &,

`For "S_2,|1-k|=2rArrk=-1,or,3.`

Evidently, `k=-1.`

Altogether, we have `4` tgts. :

`(m,c)=(4/3,-5/3)rArry=4/3x-5/3,i.e.,4x-3y-5=0.`

`(m,c)=(0,1)rArry=1,i.e., y-1=0.`

`(m,c)=(-3/4,5/4)rArry=-3/4x+5/4,or,3x+4y-5=0.`

`x=-1, or, x+1=0.`

Enjoy Maths.!