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JEE Mathematics 2024 Question with Solution

Suppose the solution of the differential equation dydx=[(2+α)x−βy+2][βx−2αy−(βy−4α)]\frac{dy}{dx} = \frac{[(2 + \alpha)x - \beta y + 2]}{[\beta x - 2\alpha y - (\beta y - 4\alpha)]} represents a circle passing through origin. Then the radius of this circle is:

  • A

    17\sqrt{17}

  • B

    12\frac{1}{2}

  • C

    172\frac{\sqrt{17}}{2}

  • D

    22

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The differential equation is

dydx=(2+α)x−βy+2βx−2αy−(βγ−4α)\frac{dy}{dx} = \frac{(2 + \alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta \gamma - 4\alpha)}

and its solution represents a circle passing through the origin.

Find: The radius of the circle.

Rewrite it in the form

M(x,y) dx+N(x,y) dy=0M(x,y) \, dx + N(x,y) \, dy = 0

by cross-multiplication:

(βx−2αy−(βγ−4α)) dy=((2+α)x−βy+2) dx(\beta x - 2\alpha y - (\beta \gamma - 4\alpha)) \, dy = ((2 + \alpha)x - \beta y + 2) \, dx

So,

((2+α)x−βy+2) dx−(βx−2αy−(βγ−4α)) dy=0((2 + \alpha)x - \beta y + 2) \, dx - (\beta x - 2\alpha y - (\beta \gamma - 4\alpha)) \, dy = 0

Here,

M=(2+α)x−βy+2,M = (2 + \alpha)x - \beta y + 2, N=−βx+2αy+(βγ−4α)N = -\beta x + 2\alpha y + (\beta \gamma - 4\alpha)

Now check exactness:

∂M∂y=−β,∂N∂x=−β\frac{\partial M}{\partial y} = -\beta, \qquad \frac{\partial N}{\partial x} = -\beta

Hence the equation is exact.

Integrating,

∫M dx+∫(terms in N not containing x) dy=C\int M \, dx + \int (\text{terms in } N \text{ not containing } x) \, dy = C

we get

(2+α)x22−βxy+2x+2αy22+(βγ−4α)y=C\frac{(2 + \alpha)x^2}{2} - \beta xy + 2x + \frac{2\alpha y^2}{2} + (\beta \gamma - 4\alpha)y = C

Multiplying by 22,

(2+α)x2+2αy2−2βxy+4x+2(βγ−4α)y−C′=0(2 + \alpha)x^2 + 2\alpha y^2 - 2\beta xy + 4x + 2(\beta \gamma - 4\alpha)y - C' = 0

For this second-degree equation to represent a circle:

  1. The coefficient of xyxy must be zero.
  2. The coefficients of x2x^2 and y2y^2 must be equal.

Thus,

−2β=0  ⟹  β=0-2\beta = 0 \implies \beta = 0

and

2+α=2α  ⟹  α=22 + \alpha = 2\alpha \implies \alpha = 2

Substitute α=2\alpha = 2 and β=0\beta = 0:

4x2+4y2+4x−16y−C′=04x^2 + 4y^2 + 4x - 16y - C' = 0

Since the circle passes through the origin, substitute x=0,y=0x = 0, y = 0:

−C′=0  ⟹  C′=0-C' = 0 \implies C' = 0

Hence,

4x2+4y2+4x−16y=04x^2 + 4y^2 + 4x - 16y = 0

Dividing by 44,

x2+y2+x−4y=0x^2 + y^2 + x - 4y = 0

Compare with the standard circle

x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

So,

2g=1  ⟹  g=12,2f=−4  ⟹  f=−2,c=02g = 1 \implies g = \frac{1}{2}, \qquad 2f = -4 \implies f = -2, \qquad c = 0

Radius is

R=g2+f2−cR = \sqrt{g^2 + f^2 - c}

Therefore,

R=(12)2+(−2)2=14+4=174=172R = \sqrt{\left(\frac{1}{2}\right)^2 + (-2)^2} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2}

The correct option is C.

Direct Circle-Condition Approach

Given: The integrated form of the exact differential equation leads to a second-degree curve.

Find: Its radius when that curve is a circle through the origin.

From the solution working, the integrated equation is

(2+α)x2+2αy2−2βxy+4x+2(βγ−4α)y−C′=0(2 + \alpha)x^2 + 2\alpha y^2 - 2\beta xy + 4x + 2(\beta \gamma - 4\alpha)y - C' = 0

For a circle, the xyxy term must vanish, so

β=0\beta = 0

Also, coefficients of x2x^2 and y2y^2 must be equal:

2+α=2α  ⟹  α=22 + \alpha = 2\alpha \implies \alpha = 2

Substituting these values gives

4x2+4y2+4x−16y−C′=04x^2 + 4y^2 + 4x - 16y - C' = 0

Since the circle passes through the origin,

C′=0C' = 0

Hence,

x2+y2+x−4y=0x^2 + y^2 + x - 4y = 0

Now use the radius formula for

x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

with

g=12,f=−2,c=0g = \frac{1}{2}, \qquad f = -2, \qquad c = 0

Therefore,

R=g2+f2−c=172R = \sqrt{g^2 + f^2 - c} = \frac{\sqrt{17}}{2}

So the correct option is C.

Common mistakes

  • Setting only the coefficient of xyxy equal to zero and forgetting that for a circle the coefficients of x2x^2 and y2y^2 must also be equal. This gives an incomplete condition. Always apply both circle conditions together.

  • Not using the fact that the circle passes through the origin. If x=0,y=0x = 0, y = 0 is not substituted, the constant term remains undetermined. After obtaining the curve equation, always impose the origin condition to find the constant.

  • Using the radius formula incorrectly by taking g=1g = 1 and f=−4f = -4 directly from x2+y2+x−4y=0x^2 + y^2 + x - 4y = 0. In the standard form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, the coefficients of xx and yy are 2g2g and 2f2f, not gg and ff.

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