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Q: What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?

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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) Point of contact: (3, 2) Slope of radius: 1/8 Slope of tangent: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26

Circle equation: x^2 +y^2 +6x -10y = 0 Completing the squares: (x +3)^2 +(y -5)^2 = 34 Center of circle: (-3, 5) Point of contact: (0, 0) Slope of radius: -5/3 Slope of tangent line: 3/5 Tangent line equation: y = 0.6x

Equation of circle: x^2 +y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Radius of circle: 17 Center of circle: (4, 8) Point of contact: (21, 8) Slope of radius: 0 Slope of tangent line: 0 Equation of tangent line: x = 21 which means it touches the circle at (21, 0) which is a straight vertical line parallel to the y axis

Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius

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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x^2 + y^2 -8x + 4y = 30 & y = x + 4 Where the tangent touches The (x,y) values will be the same. So squaring the linear equation and substituting. y^2 = (x + 4)^2 Hence x^2 + (x + 4)^2 - 8x + 4(x + 4) = 30 Multiply out x^2 + x^2 + 8x + 16 -8x + 4 x + 16 = 30 Collect 'like terms'. 2x^2 + 4x + 32 = 30 2x^2 + 4x + 2 = 0 Divide by '2' x^2 + 2x + 1 = 0 Factor (x +1)^2 = 0 x = -1 When x = -1 y = 3 This is the point of contact of the tangent with the circumference. Since x = -8x , then the x-co-ord of the centre is '4' and y = 4y , y-co-ord of the centre is '-2'. In (x,y) tangent point is (-1,3) and circle centre is (4,-2). The radius is the distance between these two points. By Pythagoras. (-1 - 4)^2 + (3 - - 2)^2 = r^2 (-5)^2 + (5)^2 = r^2 25 + 25 = r^2 50 = r^2 r = sqrt(50) = sqrt(2 x 25) = 5sqrt(2) Done !!!! Hope that helps!!!

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) Point of contact: (3, 2) Slope of radius: 1/8 Slope of tangent: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26

Circle equation: x^2 +y^2 +6x -10y = 0 Completing the squares: (x +3)^2 +(y -5)^2 = 34 Center of circle: (-3, 5) Point of contact: (0, 0) Slope of radius: -5/3 Slope of tangent line: 3/5 Tangent line equation: y = 0.6x

Equation of circle: x^2 +y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Radius of circle: 17 Center of circle: (4, 8) Point of contact: (21, 8) Slope of radius: 0 Slope of tangent line: 0 Equation of tangent line: x = 21 which means it touches the circle at (21, 0) which is a straight vertical line parallel to the y axis

Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius

Circle equation: x^2 -4x +y^2 -6y = 4 Completing the squares: (x-2)^2 +(y-3)^2 = 17 Point of contact: (6, 4) Center of circle: (2, 3) Slope of radius: 1/4 Slope of tangent line: -4 Tangent equation: y-4 = -4(x-6) => y = -4x+28 Tangent line equation in its general form: 4x+y-28 = 0

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.

Differentiate the circle equation to find the slope at any given point. x^2 + y^2 -6x + 4y -7 = 0 2x + 2y(dy/dx) - 6 + 4dy/dx =0 2x - 6 + (2y + 4) dy/dx =0 dy/dx = (6 - 2x) / (2y + +4) At the point ( 1,2) dy/dx = (6 - 2(1)) / (2(2) + 4) dy/dx = 3 / 8 The slope /gradient of the tangent line. At the point ( 1,2) y - 2 = (3/8)(x - 1) y - 2 = 3x/8 - 3/8 y = 3x/8 + 13/8 or 8y = 3x + 13 or 8y - 3x = 13 or 8y - 3x - 13 = 0 8

Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, -2) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5

Point of contact: (21, 8) Equation of circle: x^2 -y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) and its radius is 17 Slope of radius: 0 Slope of tangent: 0 Tangent equation of the circle: x = 21 meaning that the tangent line is parallel to the y axis and that the radius is parallel to the x axis.

Equation of circle: x^2 +y^2 -x -31 = 0 Completing the squares: (x-0.5)^2 +y^2 = 31.25 Center of circle: ( 0.5, 0) Point of contact: (-2, 5) Slope of radius: (0-5)/(0.5--2) = -2 Slope of tangent line: 0.5 Tangent line equation: y-5 = 0.5(x--2) => y = 0.5x+6

Circle equation: x^2 +y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) Radius of circle 17 Slope of radius: 0 Perpendicular tangent slope: 0 Tangent point of contact: (21, 8) Tangent equation: x = 21 passing through (21, 0)

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