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If: y = 2x and x^2 +y^2 -8x -y +5 = 0

Then: x^2 +(2x)^2 -8x -2x +5 = 0

So: x^2 +4x^2 -8x -2x +5 = 0 => 5x^2 -10x +5 = 0

Divide all terms by 5: x^2 -2x +1 = 0

Factorize: (x-1)(x-1) = 0 => x = 1 and x also = 1

Therefore the tangent makes contact with the circle at (1, 2)

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Q: What is the point of contact that the tangent line y equals 2x touches the circle x square plus y square -8x -y plus 5 equals 0 showing work?
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