The differential equation of the family of circles of radius 2 whose centres lie on the x-axis is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y^2\left(1 + \left(\displaystyle\frac{dy}{dx}\right)^2\right) = 4\)
A circle of radius 2 with centre (a, 0) on the x-axis is
\((x - a)^2 + y^2 = 4\)
There is one arbitrary constant a. Differentiating with respect to x:
\( \displaystyle 2(x - a) + 2y\frac{dy}{dx} = 0 \)
\( \displaystyle x - a = -y\frac{dy}{dx} \)
Eliminating a by putting this in the equation of the circle:
\( \displaystyle y^2\left(\frac{dy}{dx}\right)^2 + y^2 = 4 \)
\( \displaystyle y^2\left(1 + \left(\frac{dy}{dx}\right)^2\right) = 4 \)
Why the other options are incorrect
Option (1) \(1 + \left(\frac{dy}{dx}\right)^2 = 4y^2\) — Incorrect. The factor \(y^2\) is put on the wrong side. At the top point (a, 2) of a circle the slope is 0, and this option would give 1 = 16.
Option (2) \(x^2\left(1 + \left(\frac{dy}{dx}\right)^2\right) = 4\) — Incorrect. This is the equation for circles of radius 2 with centres on the y-axis, \(x^2 + (y - b)^2 = 4\).
Option (3) \(y^2\left(1 - \left(\frac{dy}{dx}\right)^2\right) = 4\) — Incorrect. The square of \(-y\frac{dy}{dx}\) is \(+y^2\left(\frac{dy}{dx}\right)^2\), so the sign must be plus.