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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?
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∙ 7y ago
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.
Wiki User
∙ 7y ago
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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 point of contact when the tangent line y -3x -5 equals 0 meets 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 of x^2 + y^2 -2x +4y -5 = 0 at (-2, -1)
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
Ab and ac are tangents to a circle with centre o and radius 5cmif ac equals 12cmfind the perimeter of quadrilateral aboc?
Since AB and AC are tangent to the circle O, it seems that they both are drawn from the same outside point A. As tangents to a circle from an outside point are congruent, AB ≅ BC. Also, a tangent is perpendicular to radius drawn to point of contact. So that OB and OC are congruent radii. Therefore, the perimeter of the quadrilateral ABOC equals to P = 2(12 cm) + 2(5 cm) = 34 cm.
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 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 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 solution when y equals 2x plus 1 is a tangent to the circle 5y2 plus 5x2 equals 1?
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
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 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 point of contact when the tangent line y -3x -5 equals 0 meets 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 of x^2 + y^2 -2x +4y -5 = 0 at (-2, -1)
What is the point of contact when the tangent line y equals x plus c touches the curve y equals 3x -x -5x squared?
-2
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 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 point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?
Equations: y = x+4 and x^2 +y^2 -8x +4y = 30 It appears that the given line is a tangent line to the given circle and the point of contact works out as (-1, 3)
What is the tangent equation line of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the point of contact 3 2 on the Cartesian plane?
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) Slope of radius: 1/8 Slope of tangent line: -8 Point of contact: (3, 2) Equation of tangent line: y-2 = -8(x-3) => y = -8x+26 Note that the tangent line meets the radius of the circle at right angles.
