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What is the length of the tangent line from the point 7 -2 to a point where it touches the circle of x2 plus y2 -10y -24 equals 0 on the Cartesian plane?
Equation of circle: x^2 +y^2 -10y -24 = 0
Completing the square: x^2 +(y-5)^2 = 49
Center of circle: (0, 5)
Radius of circle: 7
Distance from (7, -2) to (0, 5) is 7*square root of 2 is hypotenuse of right triangle
Using Pythagoras theorem: hypotenuse squared minus radius square = 49
Length of tangent line is the square root of 49 which is 7
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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 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 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 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 definition of a tangent line?
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
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 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 length of the tangent line from the point 0 0 to a point where it touches the circle x2 plus y2 plus 4x -6y plus 10 equals 0 on the Cartesian plane?
Equation of the circle: x^2 +y^2 +4x -6y +10 = 0 Completing the squares: (x+2)^2 +(y-3)^2 = 3 Radius of the circle: square root of 3 Center of circle: (-2, 3) Distance from (0, 0) to (-2, 5) = sq rt of 13 which is the hypotenuse of right triangle. Using Pythagoras' theorem : distance squared - radius squared = 10 Therefore length of tangent line is the square root of 10 Note that the tangent of a circle meets its radius at right angles.
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 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 length of the tangent line from the point 8 2 to a point where it touches the circle of x2 plus y2 -4x -8y -5 equals 0 on the Cartesian plane?
The distance from (8, 2) to the center of the circle forms the hypotenuse of a right angle triangle with the circle's radius meeting the tangent line at right angles and so:- Equation of the circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 = 25 Center of circle: (2, 4) Radius of circle: 5 Distance from (8, 2) to (2, 4): 2 times square root of 10 Using Pythagoras' theorem: distance squared minus radius squared = 15 Therefore length of the tangent line is the square root of 15
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
