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If: x-2y+12 = 0 then x = 2y-12 or x^2 = (2y-12)^2

If: x^2+y^2 -x -31 = 0 then (2y-12)^2 +y2 -(2y-12)-31

Expanding brackets: 4y^2 -48y +144+y^2 -2y+12-31 = 0

Collecting like terms: 5y^2 -50y+125 = 0

Divide all terms by 5: y^2 -10y+25 = 0

Factorizing the above quadratic equation: (y-5)(y-5) = 0 meaning y =5

Therefore by substitution point of contact is made at: (-2, 5) and that the line is a tangent to the circle.

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6y ago
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6y ago

The point is (-2, 5).

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Q: What is the point of contact when the line x -2y plus 12 equals 0 touches the circle x2 plus y2 -x -31 equals 0?
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