Question

The differential equation of all circles passing through the origin and having their centres on the x-axis is

$straight x squared space equals straight y squared space plus space xy dy over dx$
$straight x squared space equals straight y squared space plus space 3 xy dy over dx$
$straight x squared space equals straight y squared space plus space 2 xy dy over dx$
$straight x squared space equals straight y squared space minus space 2 xy dy over dx$

Solution

straight x squared space equals straight y squared space plus space 2 xy dy over dx

General equation of all such circles is
x²+ y² + 2gx = 0.
Differentiating, we get
$2 straight x space plus 2 straight y space dy over dx space plus 2 straight g space equals space 0 therefore space Desired space equation space is straight x squared space plus space straight y squared space plus space open parentheses negative 2 straight x minus 2 straight y dy over dx close parentheses space straight x space equals space 0 rightwards double arrow straight y squared space equals space straight x squared space plus space 2 xy dy over dx$

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For Daily Free Study Material Join wiredfaculty

Conic Section

The differential equation of all circles passing through the origin and having their centres on the x-axis is

Some More Questions From Conic Section Chapter

The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

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Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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