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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 Differential equation of circles tangent to x-axis with solution?
If the circle and x-axis are tangential then the centre of the circle must be at a distance equal to the radius away from the x-axis. That is, the centre of the circle must lie on y = ±r (where r = radius) so that the coordinates of the centre are (a, ±r) So the equation of the set of circles is (x - a)2 + (y ± r)2 = r2 Differentiating gives 2*(x - a) + 2*(y ± r)*dy/dx = 0 (x - a) + (y ± r)*dy/dx = 0 so that dy/dx = (a - x)/(y ± r) It is not clear why you would want the differential equation, but you asked for it so you've got it!
What are the tangent equations of the circle x2 plus y2 -6x plus 4x plus 5 equals 0 when it cuts through the x axis?
Equation of circle: x^2 +y^2 -6x+4y+5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, 2) The tangent lines touches the circle on the x axis at: (1, 0) and (5, 0) 1st tangent equation: y = x-1 2nd tangent equation: y = -x+5 Note that the tangent line of a circle meets its radius 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 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 Differential equation of circles tangent to x-axis?
[1+(y')2]3 = [yy"+1+(y')2]2
The center of a circle is located at c(3-13). The x-axis is tangent to the circle at (30). Find the equation of the circle?
Equation of the circle: (x-3)^2 +( y+13)^2 = 169
Find the differential equation of all circles tangent to y-axis?
Let the circle with centre (a, b) be tangent to the y-axis. Then, the radius of the circle must be b. Therefore the equation of the circle is (x - a)2 + (y - b)2 = b2 or x2 - 2ax + a2 + y2 - 2by = 0 Then 2x - 2a + 2ydy/dx - 2bdy/dx = 0 ie x - a + ydy/dx - bdy/dx (y - b)dy/dx = a - x so dy/dx = (a - x)/(y - b) or -(x - a)/(y - b)
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 Differential equation of circles tangent to x-axis with solution?
If the circle and x-axis are tangential then the centre of the circle must be at a distance equal to the radius away from the x-axis. That is, the centre of the circle must lie on y = ±r (where r = radius) so that the coordinates of the centre are (a, ±r) So the equation of the set of circles is (x - a)2 + (y ± r)2 = r2 Differentiating gives 2*(x - a) + 2*(y ± r)*dy/dx = 0 (x - a) + (y ± r)*dy/dx = 0 so that dy/dx = (a - x)/(y ± r) It is not clear why you would want the differential equation, but you asked for it so you've got it!
What are the tangent line equations of the circle x2 plus y2 -6x plus 4y plus 5 equals 0 when the circle passes through the x axis?
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, -2) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5
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 are the tangent equations of the circle x2 plus y2 -6x plus 4x plus 5 equals 0 when it cuts through the x axis?
Equation of circle: x^2 +y^2 -6x+4y+5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, 2) The tangent lines touches the circle on the x axis at: (1, 0) and (5, 0) 1st tangent equation: y = x-1 2nd tangent equation: y = -x+5 Note that the tangent line of a circle meets its radius at right angles
What are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis?
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5
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
Does always a tangent to a circle pass through the center of the circle?
the tangent will never go through the center of a cirlce. The tangent is, by definition, a line that only intersects the circle at one point. If you look down a pencil along its long axis, so that it appears to be a circle, and place your finger on top of and perpendicular to the pencil, your finger is now tangent to the circle you see.
What is tangent to the x axis?
Well, since a tangent line touches a line in one spot, the Y axis could be considered tangent to the X axis.
