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
They are +/- 5*sqrt(2)
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
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)
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
They are +/- 5*sqrt(2)
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
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.
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)
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
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)
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 +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 -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
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