x^2 + y^2 -8x + 4y = 30 &
y = x + 4
Where the tangent touches The (x,y) values will be the same.
So squaring the linear equation and substituting.
y^2 = (x + 4)^2
Hence
x^2 + (x + 4)^2 - 8x + 4(x + 4) = 30
Multiply out
x^2 + x^2 + 8x + 16 -8x + 4 x + 16 = 30
Collect 'like terms'.
2x^2 + 4x + 32 = 30
2x^2 + 4x + 2 = 0
Divide by '2'
x^2 + 2x + 1 = 0
Factor (x +1)^2 = 0
x = -1
When x = -1 y = 3 This is the point of contact of the tangent with the circumference.
Since x = -8x , then the x-co-ord of the centre is '4' and y = 4y , y-co-ord of the centre is '-2'.
In (x,y) tangent point is (-1,3) and circle centre is (4,-2). The radius is the distance between these two points.
By Pythagoras.
(-1 - 4)^2 + (3 - - 2)^2 = r^2
(-5)^2 + (5)^2 = r^2
25 + 25 = r^2
50 = r^2
r = sqrt(50) = sqrt(2 x 25) = 5sqrt(2)
Done !!!! Hope that helps!!!
Centre of circle: (4, -2)
Slope of tangent: 1
Slope of radius: -1
Radius equation: y--2 = -1(x-4) => y = -x+2
The equation for the radius which intersects the tangent is x + y = 2.
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
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 +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
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
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.
x2 + y2 = 49
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
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.
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 -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 +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 +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
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
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.
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