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https://math.answers.com/geometry/Find_the_differential_equation_of_all_circles_tangent_to_y-axis

Find the differential equation of all circles tangent to y-axis? - Answers

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)



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Find the differential equation of all circles tangent to y-axis? - Answers

https://math.answers.com/geometry/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)



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https://math.answers.com/geometry/Find_the_differential_equation_of_all_circles_tangent_to_y-axis

Find the differential equation of all circles tangent to y-axis? - Answers

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)

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      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)

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