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What is the distance to the center of the circle x2 plus y2 -2x -6y plus 5 equals 0 from the x axis at the same point when the tangent line meets the x axis from the point 3 4?
Wiki User
∙ 6y ago
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
Wiki User
∙ 6y ago
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What is the distance from a point on the x axis to the centre of the circle x2 plus y2 -2x -6y plus 5 equals 0 from the same point where a tangent line meets the circle at 3 4?
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x -1)^2 +(y -3)^2 = 5 which is radius squared Center of circle: (1, 3) Tangent line is at right angles to the radius at (3, 4) and meets the x axis at (5, 0) Distance from point (5, 0) to center of circle (1, 3) = 5 units using distance formula
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 chord that passes through the centre of a circle?
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...
What is the equation of the tangent line that touches the circle x squared plus y squared -8x -16y -209 equals 0 at a coordinate of 21 and 8?
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)
He tangent of an angle equals the ratio of the?
opposite/ adjacent
What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?
x2 + y2 = 49
What is the distance from a point on the x axis to the centre of the circle x2 plus y2 -2x -6y plus 5 equals 0 from the same point where a tangent line meets the circle at 3 4?
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x -1)^2 +(y -3)^2 = 5 which is radius squared Center of circle: (1, 3) Tangent line is at right angles to the radius at (3, 4) and meets the x axis at (5, 0) Distance from point (5, 0) to center of circle (1, 3) = 5 units using distance formula
What is the distance from a defined point on the x axis to the centre of circle x2 plus y2 -2x -6y plus 5 equals 0 when its tangent is at 3 4 on the Cartesian plane?
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 contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) => y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula
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 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 is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?
Equation of circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 = 25 which is radius squared Center of circle: (2, 4) Tangent line originates from: (8, 2) Distance from (8, 2) to (2, 4) is sq rt of 40 which is hypotenuse of right angle triangle Using Pythagoras theorem: distance^2 minus radius^2 = 15 Therefore length of tangent line is the square root of 15
What is the length of the tangent line from the point 9 0 to the circle x2 plus 8x plus y2 -9 equals 0?
Equation of circle: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Radius of circle: 5 Center of circle: (-4, 0) Distance from (9, 0) to (-4, 0) = 13 which is the hypotenuse of a right angle triangle Using Pythagoras: 13^2 -5^2 = 144 and its square root is 12 Therefore the length of the tangent line is 12 units Note that the tangent line of a circle meets the radius of the circle 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
What is the length of the tangent line from the point 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
Equation of circle: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Radius of circle: 5 Center of circle: (-4, 0) Distance from (9, 0) to (-4, 0) is 13 which is the hypotenuse of a right angle triangle Using Pythagoras' theorem: 13^2 -5^2 = 144 and its square root is 12 Therefore length of tangent line is: 12 units Note that a tangent line always meets the radius of a circle at right angles.
What is the solution when y equals 2x plus 1 is a tangent to the circle 5y2 plus 5x2 equals 1?
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
What is the length of the tangent line from the coordinate of 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
Circle equation: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Center of circle: (-4, 0) Radius of circle: 5 Distance from (-4, 0) to (9, 0) = 13 which will be the hypotenuse of a right triangle Length of tangent line using Pythagoras; theorem: 13^2 -5^2 = 144 Therefore length of tangent line is the square root of 144 = 12 units
What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?
Circle equation: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Centre of circle: (1, 3) Tangent line meets the x-axis at: (0, 5) Distance from (0, 5) to (1, 3) = 5 units using the distance formula
