0
If: 3x -2y -26 = 0
Then: y = 1.5x -13
If: x^2 +y^2 -6x +4y = 0
Then: x^3 +(1.5x -13)^2 -6x +4(1.5x -13 = 0
Expanding brackets: x^2 +2.25x^2 -39x +169 -6x +6x -52 = 0
Collecting like terms: 3.25x^2 -39x +117 = 0
Divide all terms by 3.25: x^2 -12x +36 = 0
Factorizing: (x-6)(x-6) = 0 => x = 6 and also x = 6
By substitution point of contact is at: (6, -4)
What else can I help you with?
What is the tangent equation of the circle 2x squared plus 2y squared -8x -5y -1 that touches the point of 1 -1 on the Cartesian plane?
If you mean: 2x^2 +2y^2 -8x -5y -1 = 0 making contact at (1, -1) Then the tangent equation in its general form works out as: 4x+9y+5 = 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 does 1 - tangent squared equal?
pineapple
How do you solve secant squared divided by tangent?
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
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 tangent equation of the circle 2x squared plus 2y squared -8x -5y -1 that touches the point of 1 -1 on the Cartesian plane?
If you mean: 2x^2 +2y^2 -8x -5y -1 = 0 making contact at (1, -1) Then the tangent equation in its general form works out 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
What is the value of c when y equals x plus c is a tangent to the curve y equals 3 -x -5x squared hence finding the point of contact of the line with the curve showing work?
-2
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 does 1 - tangent squared equal?
pineapple
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
How do you solve secant squared divided by tangent?
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
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 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)
What is the point of contact when the tangent line y -3x -5 equals 0 touches the circle x squared plus y squared -2x plus 4y -5 equals 0 on the Cartesian plane showing work?
If: y -3x -5 = 0 Then: y = 3x+5 If: x^2 +y^2 -2x +4x -5 = 0 Then: x^2 +(3x+5)^2 -2x+4(3x+5)-5 = 0 Removing brackets: x^2 +9x^2 +30x +25 -2x +12x +20 -5 = 0 Collecting like terms: 10x^2 +40x +40 = 0 Divide all terms by 10: x^2 +4x +4 = 0 Factorizing: (x+2)(x+2) = 0 => x = -2 and also x = -2 Therefore by substitution the tangent line makes contact with the circle at (-2, -1) on the Cartesian plane.
What is the point of contact that the tangent line x -2y plus 12 equals 0 touches the circle x squared plus y squared -x -31 equals 0 on the Catesian plane showing work?
If: x -2y +12 = 0 Then: x = 2y -12 If: x^2 +y^2 -x -31 = 0 Then: (2y -12)^2 +y^2 -(2y -12) -31 = 0 So: 4y^2 -48y +144 +y^2 -2y +12 -31 = 0 Collecting like terms: 5y^2 -50y +125 = 0 Using the quadratic equation formula: y = 5 and y also = 5 By substituting: x = -2 and y = 5 Therefore the tangent line makes contact with the circle at (-2, 5) on the Cartesian plane.
What is the tangent equation that touches the circle 2x squared plus 2y squared -8x -5y -1 equals 0 at the point of 1 -1 on the Cartesian plane showing work?
Circle equation: 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms by 2: x^2 +y^2 -4x -2.5y -0.5 = 0 Complete the squares: (x-2)^2 +(y-1.25)^2 -4 -1.5625 -1 = 0 So: (x-2)^2 +(y-1.25)^2 = 6.0625 Centre of circle: (2, 1.25) Contact point: (1, -1) Slope of radius: (-1-1.25)/(1-2) = 9/4 Slope of tangent line: -4/9 Tangent equation: y- -1 = -4/9(x-1) => 9y--9 = -4x+4 => 9y = -4x-5 Tangent equation in its general form: 4x+9y+5 = 0
