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What is the tangent equation of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the coordinate of 3 2?
Wiki User
∙ 6y ago
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
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∙ 6y ago
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What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?
x2 + y2 = 49
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
What is the tangent equation of the circle x2 plus 6 plus y2 -10 equals 0 when it passes through 0 0 on the Cartesian plane?
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
What is the tangent equation that touches the circle x2 -y2 -8x -16y -209 equals 0 at the point of 21 and 8 on the Cartesian plane?
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.
What is the tangent line equation of the circle x2 plus y2 -8x -16y -209 equals 0 when it touches the circle at 21 8 on the Cartesian plane?
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
What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?
x2 + y2 = 49
What is the equation of the tangent line that touches the circle x squared plus y squared -8x -16y -209 equals 0 at a coordinate of 21 and 8?
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)
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
What is the distance from a defined point on the x axis to the centre of circle x2 plus y2 -2x -6y plus 5 equals 0 when its tangent is at 3 4 on the Cartesian plane?
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
What is the tangent equation of the circle x2 plus 6 plus y2 -10 equals 0 when it passes through 0 0 on the Cartesian plane?
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
What is the tangent equation that touches the circle x2 -y2 -8x -16y -209 equals 0 at the point of 21 and 8 on the Cartesian plane?
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.
What is the tangent line equation of the circle x2 plus y2 -8x -16y -209 equals 0 when it touches the circle at 21 8 on the Cartesian plane?
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
What is the tangent line equation that passes through the origin touching the circle x2 plus y2 plus 6x -10y equals 0?
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
What is the point of contact when the tangent line y -3x -5 equals 0 touches the circle x2 plus y2 -2x plus 4y -5 equals 0?
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.
What are the equations of the tangent and normal at the coordinate of 6 4 of the circle x2 -4x plus y2 -6y equals 4 when plotted on the Cartesian plane?
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
What is the tangent equation at the point of 1 -1 when it touches the circle of 2x2 plus 2y2 -8x -5y -1 equals 0?
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
What is the tangent equation that touches the circle of x squared plus y squared -8x -y plus 5 equals 0 at the point of 1 2 on the Cartesian plane showing work?
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
