0
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 else can I help you with?
How do you determine the length of the radius on a circle?
The radius of a circle is defined as the distance from the centre-point to the circumference.
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?
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
What is the centre and radius of a circle enscribed through the points of 6 2 and 8 -2 and 0 2 on the Cartesian plane?
It works out that the circle's centre is at (3, -2) and its radius is 5 on the Cartesian plane.
What is the distance from the centre of Wodonga to the centre of Albury?
From the centre of Albury to the centre of Wodonga is a distance of 5.53km.
What is the Cartesian equation of a circle whose centre is in the 1st quadrant with a radius of 7 and touches the x and y axes?
Centre of the circle is at (7, 7) and its Cartesian equation is (x-7)^2 + (y-7)^2 = 49
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
What is the name of a segment that doesn't have a vertices's in the centre of the circle?
You maybe referring to the diameter of a circle or its tangent
What is the equation of a circle whose centre is at 3 -5 and meets the point of 6 -7 on the Cartesian plane?
Centre of circle: (3, -5) Distance from (3, -5) to (6, -7) is the square root of 13 which is the radius Equation of the circle: (x-3)^2 + (y+5)^2 = 13
What is a tangent line with respect to a circle?
A tangent to a circle is a line which touches the circle once. That is, it does not pass through the circle, which would mean intersecting it twice. A way to form a tangent is draw any line from the centre point of a circle to its edge. A line on the edge perpendicular (at 90 degrees to) this line will be a tangent.
Distance the moon is from the earth?
The average centre-to-centre distance from the Earth to the Moon is 384,403 km
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 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.
