Question:medium

The differential equation of the family of all circles of radius 'a' is

Show Hint

The differential equation of a family of curves can often be found from a defining geometric property. For a family of circles with a constant radius, the key property is that their radius of curvature is constant and equal to the radius of the circle.

Updated On: Mar 30, 2026

$y_1y_2 + (1+y_1^2)=a$
$(1+y_1^2)^3 = a^2y_2^2$
$1+y_1^2 = y_2^2+a^2$
$y_2^2+1 = y_1^2+a^2$

Show Solution

The Correct Option is B

Solution and Explanation

Was this answer helpful?

0

Top Questions on Differential equations

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra