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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

Q: What is the tangent equation of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the coordinate of 3 2?

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x2 + y2 = 49

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 passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x

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 -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

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x2 + y2 = 49

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)

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 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) => y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula

Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x

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 -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 +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

It works out that the tangent line of y -3x -5 = 0 makes contact with the circle x^2 +y^2 -2x +4y -5 = 0 at the coordinate of (-2, -1) on the coordinated grid.

Circle equation: x^2 -4x +y^2 -6y = 4 Completing the squares: (x-2)^2 +(y-3)^2 = 17 Center of circle: (2, 3) Point of contact: (6, 4) Slope of radius: 1/4 Slope of tangent: -4 Slope of normal: 1/4 Equation of tangent: y-4 = -4(x-6) => y = -4x+28 Equation of normal: y-4 = 1/4(x-6) => 4y-16 = x-6 => 4y = x+10

The tangent equation that touches the circle 2x^2 +2y^2 -8x -5y -1 = 0 at the point of (1, -1) works out in its general form as: 4x +9y +5 = 0

Equation of circle: x^2 +y^2 -8x -y +5 = 0Completing the squares: (x-4)^2 +(y-0.5)^2 = 11.25Centre of circle: (4, 0.5)Slope of radius: -1/2Slope of tangent: 2Equation of tangent: y-2 = 2(x-1) => y = 2xNote that the above proves the tangent of a circle is always at right angles to its radius