0
What are the lengths of the tangent lines when they touch the circle x squared plus y squared -10x plus 8y plus 5 equals 0 from the point 5 4 on the Cartesian plane?
Wiki User
∙ 7y ago
A circle with centre (x0, y0) and radius r has the formula:
(x - x0)² + (y - y0)² = r²
Completing the squares:
x² + y² - 10x + 8y + 5 = 0
→ x² -10x + 25 - 25 + y² + 8y + 16 - 16 + 5 = 0
→ (x - 5)² - 25 + (y + 4)² - 16 + 5 = 0
(x - 5)² + (y + 4)² = 36 = 6²
→ The circle has centre (5, -4) and radius 6.
A tangent to the circle forms a right angle with the radius of the circle that meets the tangent.
Joining the centre of the circle to the point (5, 4) will form the hypotenuse of the triangle with the radius and the tangent as the other two sides.
The length of the hypotenuse can be calculated using Pythagoras:
hypotenuse² = (5 - 5)² + (-4 - 4)² = 0 + 8² = 8²
Thus the length of the tangents from the point (5, 4) can be calculated using Pythagoras:
radius² + tangent² = hypotenuse²
→ 6² + tangent² = 8²
→ tangent² = 8² - 6² = 64 - 36 = 28
→ tangent = √28 = √(4×7) = √4 √7 = 2 √7
Each tangent is 2 √7 units long.
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoThey are 2*sqrt(7) units long.
Add your answer:
Earn +20 pts
What is the tangent equation of the circle 2x squared plus 2y squared -8x -5y -1 that touches the point of 1 -1 on the Cartesian plane?
If you mean: 2x^2 +2y^2 -8x -5y -1 = 0 making contact at (1, -1) Then the tangent equation in its general form works out as: 4x+9y+5 = 0
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
What are the lengths of the tangent lines from the point 8 2 when they touch the circle x2 plus y2 -4x -8y -5 equals 0 on the Cartesian plane showing work?
Tangent lines stem from the point: (8,2) Equation of circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 -4 -16 -5 = 0 So: (x-2)^2 +(y-4)^2 = 25 which is the radius squared Centre of circle: (2, 4) Distance from (2, 4) to (8, 2) = 40 which is the distance squared Lengths of tangents using Pythagoras: 40-25 = 15 => square root of 15 Note that the distance from (2, 4) to (8, 2) is actually the hypotenuse of a right angle triangle.
What are the lengths of the tangent lines from the point 7 -2 to the circle x2 plus y2 -10y -24 equals 0 on the Cartesian plane?
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) = sq rt of 98 and is the hypotenuse of a right triangle Using Pythagoras: theorem: distance^2 minus radius^2 = 49 Therefore lengths of tangent lines are square root of 49 = 7 units
Tangent of a circle is what?
A tangent of a circle is a straight line that touches the circle at only one point.
What is the tangent equation that touches the circle of x squared plus y squared -8x -y plus 5 equals 0 at the point of 1 2 on the Cartesian plane showing work?
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
What is the tangent equation of the circle 2x squared plus 2y squared -8x -5y -1 that touches the point of 1 -1 on the Cartesian plane?
If you mean: 2x^2 +2y^2 -8x -5y -1 = 0 making contact at (1, -1) Then the tangent equation in its general form works out as: 4x+9y+5 = 0
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
What are the lengths of the tangent lines from the point 8 2 when they touch the circle x2 plus y2 -4x -8y -5 equals 0 on the Cartesian plane showing work?
Tangent lines stem from the point: (8,2) Equation of circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 -4 -16 -5 = 0 So: (x-2)^2 +(y-4)^2 = 25 which is the radius squared Centre of circle: (2, 4) Distance from (2, 4) to (8, 2) = 40 which is the distance squared Lengths of tangents using Pythagoras: 40-25 = 15 => square root of 15 Note that the distance from (2, 4) to (8, 2) is actually the hypotenuse of a right angle triangle.
What is the length of the tangent line from the point 0 0 to a point where it touches the circle x2 plus y2 plus 4x -6y plus 10 equals 0 on the Cartesian plane?
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.
What are the lengths of the tangent lines from the point 7 -2 to the circle x2 plus y2 -10y -24 equals 0 on the Cartesian plane?
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) = sq rt of 98 and is the hypotenuse of a right triangle Using Pythagoras: theorem: distance^2 minus radius^2 = 49 Therefore lengths of tangent lines are square root of 49 = 7 units
Tangent of a circle is what?
A tangent of a circle is a straight line that touches the circle at only one point.
What is the Tangent Line to Circle Theorem?
The Tangent Line to Circle Theorem states that a line is tangent to a circle if and only if it's perpendicular to the circle's radius.
What is the length of the tangent line from the point 8 2 to a point where it touches the circle of x2 plus y2 -4x -8y -5 equals 0 on the Cartesian plane?
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
What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?
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
What is the equation of the tangent line that touches the circle x squared plus y squared -6x plus 4y equals 0 at the point of 6 -4 on the Cartesian plane?
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
What does the radius-tangent theorem state?
The radius-tangent theorem is math involving a circle. The radius-tangent theorem states that a line is tangent to a circle if it is perpendicular to the radius of a circle.
