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`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...
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The point of contact of two circles an be obtained by solving the two equations.
`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 x+y=2 ` --- (iii)
so, `x-y=0 ` --- (iv)
Solving eqn.s (iii) and (iv),
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)`
Therefore, the two circles touch internally, and at point P(1, 1).
Posted by llltkl on August 5, 2013 at 10:53 AM (Answer #1)
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