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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 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 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 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.
