# show that the circles are tangent to each other x^2+y^2-6x+1=0 x^2+y^2-2y+8x-1=0.analytic geometry

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### 1 Answer

The circles are tangent to each other if they touch each other only at a single point.

We have the equation of the circles:

x^2 + y^2 - 6x + 1 = 0 ...(1)

x^2 + y^2 - 2y + 8x - 1 = 0 ...(2)

Equating (1) and (2)

x^2 + y^2 - 6x + 1 = x^2 + y^2 - 2y + 8x - 1

=> - 6x + 1 + 2y - 8x + 1 = 0

=> -14x + 2y + 2 = 0

=> -7x + y + 1 = 0

=> y = -1 + 7x

substitute this in (1)

x^2 + (-1 + 7x)^2 - 6x + 1 = 0

=> x^2 + 1 + 49x^2 - 14x - 6x + 1 = 0

=> 50x^2 - 20x + 2 = 0

=> 50x^2 - 10x - 10x +2 =0

=> 10x(5x - 1) - 2(5x - 1) = 0

=> (10x - 2)(5x - 1) = 0

=> x = 2/10 = 1/5 and x = 1/5

y = -1 + 7x = -1 + 7/5 = 2/5

The two circles meet only at the point (1/5 , 2/5)

**This proves that they are tangent to each other**