# Find the point of contact of the curves defined by x^2 + y^2 - 3x - 2y = 6 and 3x^2 + 2y^2 - 2x - 4y = 4.

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### 1 Answer

The point of contact of the curves defined by x^2 + y^2 - 3x - 2y = 6 and 3x^2 + 2y^2 - 2x - 4y = 4 has to be determined.

x^2 + y^2 - 3x - 2y = 6 ...(1)

3x^2 + 2y^2 - 2x - 4y = 4 ...(2)

(2) - 2*(1)

=> 3x^2 - 2x^2 + 2y^2 - 2y^2 - 2x + 6x - 4y + 4y = 4 - 12

=> x^2 + 4x + 8 = 0

=> `x = (-4+-sqrt(16-32))/2`

This quadratic equation does not have any real roots.

**The two curves do not intersect each other.**