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Equation of the circle: x^2 +y^2 -2x +4y -5 = 0

Completing the squares: (x-1)^2 +(y+2)^2 = 10

Centre of circle: (1, -2)

Slope of the radius: -1/3 because it is perpendicular to the tangent line

Equation of the radius: y --2 = -1/3(x-1) => 3y = -x-5

Solving the simultaneous equations of: 3y = -x-5 and y = 3x+5 => x = -2, y = -1

Therefore the point of contact is at: (-2, -1)

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Q: What is the point of contact when the tangent line y equals 3x plus 5 touches the circle x squared plus y squared -2x plus 4y -5 equals 0 on the Cartesian plane?
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