`S1=x^2+y^2-2=0`  and `S2=x^2+y^2+3x+3y-8=0` Show that S1=0 and S­2=0 touch internally , and find the coordinates of  the point of contact P.



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The point of contact of two circles an be obtained by solving the two equations.

Here, `S_1=x^2+y^2-2=0` 

`rArr x^2+y^2=2` ---- (i)

Putting this value in `S_2` , i.e. `x^2+y^2+3x+3y-8=0`

`rArr x^2+y^2+3x+3y=8` --- (ii)

`rArr 2+3x+3y-8=0`

`rArr x+y=2 ` --- (iii)

Upon squaring,


`rArr 2+2xy=4`


Therefore, `x-y=sqrt((x+y)^2-4xy)`



so, `x-y=0 ` --- (iv)

Solving eqn.s (iii) and (iv),

`x=1, y=1`

Hence, the coordinates of the point of contact P, is (1, 1).

Two circles touch internally at a point P if the distance between their centres is equal to difference of their radii, i.e.

`c_1.c_2=(r_2-r_1) or (r_1-r_2)`

The equations of `S_1` and `S_2` can be rearranged as:



So, `S_1` has its centre, `c_1` at (0, 0), radius, `r_1=sqrt2`

and `S_2` has its centre, `c_2` at (-3/2, -3/2), radius, `r_2=5/sqrt2`

Distance between `c_1` and `c_2` `=sqrt((-3/2-0)^2+(-3/2-0)^2)`

`=2.12132` units


`=2.12132` units

Therefore, the two circles touch internally, and at point          P(1, 1).


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