The distance from (8, 2) to the center of the circle forms the hypotenuse of a right angle triangle with the circle's radius meeting the tangent line at right angles and so:-
Equation of the circle: x^2 +y^2 -4x -8y -5 = 0
Completing the squares: (x-2)^2 +(y-4)^2 = 25
Center of circle: (2, 4)
Radius of circle: 5
Distance from (8, 2) to (2, 4): 2 Times Square root of 10
Using Pythagoras' theorem: distance squared minus radius squared = 15
Therefore length of the tangent line is the square root of 15
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)
opposite/ adjacent
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 line from (3, 4) meets the x axis at: (5, 0) Distance from (5, 0) to (1, 3) = 5 using the distance formula
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...
The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (0, 0), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (0, 0) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (0, 0) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r.Use Pythagoras to find the length between the point (0, 0) and the centre of the circle (Xo, Yo)Use Pythagoras to find the length between the point (0, 0) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² + 4x - 6y + 10 = 0→ x² + 4x + y² - 6y + 10 = 0→ (x + (4/2))² - (4/2)² + (y + (-6/2))² - (-6/2)² +10 = 0→ (x + 2)² - 4 + (y - 3)² - 9 + 10 = 0→ (x + 2)² + (y - 3)² = 3 = radius²→ Circle has centre (-2, 3) and radius √3Line from centre of circle (-2, 3) to the given point (0, 0):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (0, 0) and centre of circle (-2, 3)→ length = √((0 - -2)² + (0 - 3)²)= √(2² + (-3)²)= √13Tangent line segment:Using Pythagoras to find length of tangent between point (0, 0) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√13)² + (√3)²)= √(13 + 3)= √16= 4
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
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
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.
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
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.
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.
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
Equation of the circle: x^2 +y^2 +4x -6y +10 = 0 Completing the squares: (x+2)^2 +(y-3)^2 = 3 Radius of the circle: square root of 3 Center of circle: (-2, 3) Distance from (0, 0) to (-2, 5) = sq rt of 13 which is the hypotenuse of right triangle. Using Pythagoras' theorem : distance squared - radius squared = 10 Therefore length of tangent line is the square root of 10 Note that the tangent of a circle meets its radius at right angles.
Equation of circle: x^2 +y^2 -10y -24 = 0 Completing the square: x^2 +(y-5)^2 = 49 Center of circle: (0, 5) Radius of circle: 7 Distance from (7, -2) to (0, 5) is 7*square root of 2 is hypotenuse of right triangle Using Pythagoras theorem: hypotenuse squared minus radius square = 49 Length of tangent line is the square root of 49 which is 7
Equation: x² + y² -6x +4y = 0 Completing the squares: (x-3)² + (y+2)² = 13 Centre of circle: (3, -2) Contact point: (6, -4) Slope of radius: -2/3 Slope of tangent: 3/2 Tangent equation: y - -4 = 3/2(x-6) => 2y - -8 = 3x-18 => 2y = 3x-26 Tangent line equation in its general form: 3x-2y-26 = 0
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)
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