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If you mean points of (3, 9) and (-4, 2) then the slope of the line is 1

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Q: What is the slope of the line shown below (39) and (-42)?
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A line with slope m and a point (x0, y0) on it has equation: y - y0 = m(x - x0) The slope of the tangent is perpendicular to the slope of the radius to the point (3, 2) The product of the slope of a line and a line perpendicular to it is -1. A circle with centre (X, Y) and radius r has equation: (x - X)² + (y - Y)² = r² For x² + 10x + y² - 2y - 39 = 0 and the point (3, 2): x² + 10x + y² - 2y - 39 = 0 → x² + 10x + (10/2)² - (10/2)² + y² - 2y + (2/2)² - (2/2)² - 39 = 0 → (x + 5)² - 25 + (y - 1)² - 1 - 39 = 0 → (x - -5)² + (y - 1)² = 65 → the circle has centre (-5, 1) and radius √65. The slope m' of the radius to (3, 2) from the centre of (-5, 1) is given by: slope = change_in_y / change_in_x → m' = (2 - 1) / (3 - -5) = 1/8 → slope m of the tangent is: mm' = -1 → m = -1/m → m = -1/(1/8) = -8 Thus the tangent has equation: y - 2 = -8(x - 3) → y - 2 = -8x + 24 → y + 8x = 26


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Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 +(y-1)^2 = 65 Center of circle: (-5, 1) Slope of radius: 1/8 Slope of tangent line: -8 Point of contact: (3, 2) Equation of tangent line: y-2 = -8(x-3) => y = -8x+26 Note that the tangent line meets the radius of the circle at right angles.


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First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0


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