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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

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It is 4x + 9y - 5 = 0

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Q: 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?
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What are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis?

Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5


What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?

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


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 is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?

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


What is the equation of the circle that touches the x axis when its center is at 2 5?

Center of circle: (2, 5) Point of contact with the x axis: (2, 0) Distance from (2, 5) to (2, 0) equals 5 which is the radius of the circle Equation of the circle: (x-2)^2 +(y-5)^2 = 25

Related questions

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 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 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


What is the equation of the tangent line that touches the circle x squared plus y squared -8x -16y -209 equals 0 at the point of 21 and 8?

Circle equation: x^2 +y^2 -8x -16y -209 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) Radius: 17 Slope of radius: 0 Tangent equation line: 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 are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis?

Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5


What are the tangent equations of the circle x2 plus y2 -6x plus 4x plus 5 equals 0 when it cuts through the x axis?

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) The tangent lines touches the circle on the x axis at: (1, 0) and (5, 0) 1st tangent equation: y = x-1 2nd tangent equation: y = -x+5 Note that the tangent line of a circle meets its radius at right angles


What is the tangent equation of the circle 2x2 plus 2y2 -8x -5y -1 equals 0 that touches the circle at the point of 1 and -1?

Point of contact: (1, -1) Equation of circle: 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms by 2: x^2 +y^2 -4x -2.5y -0.5 = 0 Completing the squares: (x-2)^2 +(y-1.25)^2 = 6.0625 Center of circle: (2, 1.25) Slope of radius: 9/4 Slope of tangent line: -4/9 Equation of tangent: y--1 = -4/9(x-1) => 9y--9 = -4x+4 => 9y = -4x-5 Equation of tangent in its general form: 4x+9y+5 = 0


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


What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?

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