Question:
If the differential equation representing the family of all circles touching \(x\)-axis at the origin is \((x^2 - y^2)\frac{dy}{dx} = g(x)y\), then \(g(x)\) equals
If the differential equation representing the family of all circles touching \(x\)-axis at the origin is \((x^2 - y^2)\frac{dy}{dx} = g(x)y\), then \(g(x)\) equals
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The general circle touching the x-axis at the origin is \(x^2 + y^2 + 2ky = 0\); eliminate \(k\) by differentiating.
Updated On: Nov 11, 2026
- \(\frac12 x^2\)
- \(2x\)
- \(\frac12 x\)
- \(2x^2\)
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The Correct Option is B
Solution and Explanation
Circles touching the \(x\)-axis at the origin: \(x^2 + y^2 + 2ky = 0\) (centre \((0,-k)\)).
Differentiate: \(2x + 2yy' + 2ky' = 0\), so \(k = -\frac{x + yy'}{y'}\).
Substitute: \(x^2 + y^2 - \frac{2y(x + yy')}{y'} = 0\), i.e. \((x^2 + y^2)y' = 2xy + 2y^2y'\).
So \((x^2 - y^2)y' = 2xy\), hence \(g(x) = 2x\).
Answer: (2) \(2x\)
Differentiate: \(2x + 2yy' + 2ky' = 0\), so \(k = -\frac{x + yy'}{y'}\).
Substitute: \(x^2 + y^2 - \frac{2y(x + yy')}{y'} = 0\), i.e. \((x^2 + y^2)y' = 2xy + 2y^2y'\).
So \((x^2 - y^2)y' = 2xy\), hence \(g(x) = 2x\).
Answer: (2) \(2x\)
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