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
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)
x2 + y2 = 49
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
Circle equation: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Center of circle: (3, -2) Radius of circle: square root of 8 The radius of the circle will touch the points of (1, 0) and (5, 0) on the x axis The tangent slope at (1, 0) is 1 The tangent slope at (5, 0) is -1 Equations of the tangents are: y = x-1 and y = -x+5
Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x
This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
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
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)
x2 + y2 = 49
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
The chord that passes through the center of a circle is called the diameter. The measure of this chord is also called the diameter, and is used in calculations such as finding the circumference of a circle (Circumference equals pi times the diameter).^^wrong...it is eitherSecant or tangent im pretty sure its SECANT though...
Circle equation: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Center of circle: (3, -2) Radius of circle: square root of 8 The radius of the circle will touch the points of (1, 0) and (5, 0) on the x axis The tangent slope at (1, 0) is 1 The tangent slope at (5, 0) is -1 Equations of the tangents are: y = x-1 and y = -x+5
Cotangent 32 equals tangent 0.031
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)
Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x
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
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