If the solution of dydx=ax+32y+⨍ represents a circle, then the value of a is

Given dydx= ax+32y+f⇒ ∫(2y+f) dy=∫(ax+3) dx⇒ 2(y22)+f(y)=a(x22)+3x+c⇒ a2x2−y2+3x−fy+c=0If represents a circle iff a2=−1⇒ a= −2

If the solution of dydx=ax+32y+⨍ represents a circle, then the value of a is

Given dydx= ax+32y+f⇒ ∫(2y+f) dy=∫(ax+3) dx⇒ 2(y22)+f(y)=a(x22)+3x+c⇒ a2x2−y2+3x−fy+c=0If represents a circle iff a2=−1⇒ a= −2

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If the solution of $\frac{dy}{dx} = \frac{ax + 3}{2 y + ⨍}$ represents a circle, then the value of a is

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–4

–2

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{d y}{d x} = \frac{a x + 3}{2 y + f} \\ \Rightarrow \int (2 y + f) d y = \int (a x + 3) d x \\ \Rightarrow 2 (\frac{y^{2}}{2}) + f (y) = a (\frac{x^{2}}{2}) + 3 x + c \\ \Rightarrow \frac{a}{2} x^{2} - y^{2} + 3 x - f y + c = 0 \\ I f r e p r e s e n t s a c i r c l e i f f \frac{a}{2} = - 1 \Rightarrow a = - 2 \end{array}$