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Q: What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?

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Centre of circle: (4, -2) Slope of tangent: 1 Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2

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) Point of contact: (3, 2) Slope of radius: 1/8 Slope of tangent: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26

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

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

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

The tangent line equation touching the given circle works out as 2y = x+3 or as x-2y+3 = 0 in its general form

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

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.

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.

56

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) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5

Equation of circle: x^2 +y^2 -x -31 = 0 Completing the squares: (x-0.5)^2 +y^2 = 31.25 Center of circle: ( 0.5, 0) Point of contact: (-2, 5) Slope of radius: (0-5)/(0.5--2) = -2 Slope of tangent line: 0.5 Tangent line equation: y-5 = 0.5(x--2) => y = 0.5x+6

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)

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

That's the equation of a circle with its center at the origin and a radius of 8.

(-4,3)

It is the equation of a circle with radius of 6 and its center at the origin.

Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) => y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula

Centre = (0,0), the origin; radius = 11

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

The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).

The question is suppose to read: Find the equation of the line tangent to y=(x²+3x)²(2x-2)³, when x=8

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

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

At the center, (x, y) = (-2, 5)