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If: 3x -2y -26 = 0

Then: y = 1.5x -13

If: x^2 +y^2 -6x +4y = 0

Then: x^3 +(1.5x -13)^2 -6x +4(1.5x -13 = 0

Expanding brackets: x^2 +2.25x^2 -39x +169 -6x +6x -52 = 0

Collecting like terms: 3.25x^2 -39x +117 = 0

Divide all terms by 3.25: x^2 -12x +36 = 0

Factorizing: (x-6)(x-6) = 0 => x = 6 and also x = 6

By substitution point of contact is at: (6, -4)

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Q: What is the point of contact when the tangent line 3x -2y -26 equals 0 touches the circle x squared plus y squared -6x plus 4y equals 0 on the Cartesian plane showing work?
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What is the equation of the tangent line that touches the circle x squared plus y squared -6x plus 4y equals 0 at the point of 6 -4 on the Cartesian plane?

Equation: x² + y² -6x +4y = 0 Completing the squares: (x-3)² + (y+2)² = 13 Centre of circle: (3, -2) Contact point: (6, -4) Slope of radius: -2/3 Slope of tangent: 3/2 Tangent equation: y - -4 = 3/2(x-6) => 2y - -8 = 3x-18 => 2y = 3x-26 Tangent line equation in its general form: 3x-2y-26 = 0