If the order of the differential equation of the family of circles touching the x -axis at the origin is k , then 2k is equal to

Question

If the order of the differential equation of the family of circles touching the x -axis at the origin is k , then 2k is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-8a1jxmn9gmc3.png" /> The equation of the family is x2+(y - r)2=r2 i.e. only one arbitrary constant. Hence, the order of its differential equation will be '1' ∴ 2k=2

Q. If the order of the differential equation of the family of circles touching the $x$ -axis at the origin is $k$ , then $2 k$ is equal to

The equation of the family is $x^{2} + (y - r)^{2} = r^{2}$
i.e. only one arbitrary constant.
Hence, the order of its differential equation will be $^{'} 1^{'}$
$∴ 2 k = 2$

Q. If the order of the differential equation of the family of circles touching the x -axis at the origin is k , then 2k is equal to

The equation of the family is x2+(y−r)2=r2 i.e. only one arbitrary constant. Hence, the order of its differential equation will be ′1′ ∴2k=2

Q. If the order of the differential equation of the family of circles touching the $x$ -axis at the origin is $k$ , then $2 k$ is equal to

The equation of the family is $x^{2} + (y - r)^{2} = r^{2}$
i.e. only one arbitrary constant.
Hence, the order of its differential equation will be $^{'} 1^{'}$
$∴ 2 k = 2$