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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
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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 is the tangent equation that touches the circle x2 -y2 -8x -16y -209 equals 0 at the point of 21 and 8 on the Cartesian plane?
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.
What is the tangent equation line of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the point of contact 3 2 on the Cartesian plane?
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.
What is the tangent line equation of the circle x2 plus y2 -8x -16y -209 equals 0 when it touches the circle at 21 8 on the Cartesian plane?
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
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 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 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 that touches the circle x2 -y2 -8x -16y -209 equals 0 at the point of 21 and 8 on the Cartesian plane?
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.
What is the tangent equation line of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the point of contact 3 2 on the Cartesian plane?
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.
What is the tangent line equation of the circle x2 plus y2 -8x -16y -209 equals 0 when it touches the circle at 21 8 on the Cartesian plane?
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
What are the lengths of the tangent lines when they touch the circle x2 -4x plus y2 plus 2y -11 equals 0 from the coordinate of 8 5 on the Cartesian plane?
The two tangents from a point to a circle are equal in length from the point to where they touch the circle. Completing the squares in x and y for the equation of the circle gives: x² - 4x + y² + 2y - 11 = 0 → (x - 4/2)² - (4/2)² + (y + 2/2)² - (2/2)² - 11 = 0 → (x - 2)² - 4 + (y + 1)² - 1 - 11 = 0 → (x - 2)² + (y + 1)² = 16 = radius² (= 4²) → The circle has centre (2, -1) and radius 4 The distance from the point to the centre of the circle forms the hypotenuse if a triangle with the radius as one leg and the tangent as the other leg (as the tangent and the radius are perpendicular). The length of the hypotenuse is calculated using Pythagoras from the Cartesian coordinates of the points at each end of it. → length_tangent = √(hypotenuse² - radius²) = √((√((8 - 2)² + (5 - -1)²))² - 16) = √(6² + 6² - 16) = √56 units = 2√14 units ≈ 7.48 units
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 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 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 tangent line equation when it makes contact with the circle x2 plus y2 -x -31 equals 0 at -2 5 on the Cartesian plane?
Equation of circle: x^2 +y^2 -x -31 = 0 Completing the squares: (x-0.5)^2 +y^2 = 31.25 Center of circle: ( 0.5, 0) Point of contact: (-2, 5) Slope of radius: (0-5)/(0.5--2) = -2 Slope of tangent line: 0.5 Tangent line equation: y-5 = 0.5(x--2) => y = 0.5x+6
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.
What is the tangent equation of the circle x2 plus 6 plus y2 -10 equals 0 when it passes through 0 0 on the Cartesian plane?
Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x
