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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?
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.
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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.
