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What is the value of k when y equals 3x plus 1 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
What are the possible value of k when the line y equals x plus k is a tangent to the circle x squared plus y squared equals 25?
They are +/- 5*sqrt(2)
How do you find the length of a leg on a right triangle when one leg and the slope of the hypotenuse is given?
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.
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 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 value of k when y equals 3x plus 1 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
At what point is the line of y equals x -4 tangent to the curve of x squared plus y squared equals 8?
y = x - 4 x^(2) + y^(2) = 8 Hence y^(2) = 8 - x^(2) Square up the linear eq'n and substitute Hence y^(2) = ( x - 4)^(2) NB Notice the RHS is wholly squared , not individual terms. Substitute ; this eliminates 'y' ( x - 4)^(2) = 8 - x^(2) Square out the brackets x^(2) - 8x + 16 = 8 - x^(2) Collect terms 2x^(2) -8x +8 = 0 Factor 2(x^(2) - 4x + 4) = 0 (x- 2)(x - 2) = 0 Hence x = 2 ( The 'x' coordinate) . Substitute into either eq;n for 'y' y = 2 - 4 y = -2 Hence ( x,y) = ( 2,-2) ; The point of contact of the tangent and the curve.
What are the possible value of k when the line y equals x plus k is a tangent to the circle x squared plus y squared equals 25?
They are +/- 5*sqrt(2)
What is the value of k when y equals 3x plus 1 is a line tangent to the curve of x squared plus y squared equals k hence finding the point of contact of the line to the curve?
It is (-0.3, 0.1)
How do you find the length of a leg on a right triangle when one leg and the slope of the hypotenuse is given?
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.
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 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
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 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 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 slope and y-intercept of y equals -2x plus 5?
y = -2x + 5 Slope is -2, that is the angle the line makes with the x-axis is such that tangent of that angle is -2 The y-intercept is 5
