Question:
The order of the differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is
The order of the differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is
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The order of the differential equation of a family of curves is usually equal to the number of arbitrary constants in the family.
Updated On: May 7, 2026
- \(4\)
- \(3\)
- \(2\)
- \(1\)
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The Correct Option is D
Solution and Explanation
Concept:
The order of a differential equation is equal to the highest order derivative present in it. For a family of curves, the order is generally equal to the number of arbitrary constants.
Step 1: Since the centres are on the \(x\)-axis, let the centre be: \[ (a,0) \]
Step 2: Since the circle passes through the origin, its radius is: \[ r=a \]
Step 3: Equation of the circle is: \[ (x-a)^2+y^2=a^2 \] Expanding: \[ x^2-2ax+a^2+y^2=a^2 \] \[ x^2+y^2-2ax=0 \]
Step 4: This family contains only one arbitrary constant \(a\).
So we need to differentiate once to eliminate \(a\).
Step 5: Differentiate: \[ x^2+y^2-2ax=0 \] \[ 2x+2y\frac{dy}{dx}-2a=0 \] The highest derivative present is: \[ \frac{dy}{dx} \] So the order is: \[ 1 \] Therefore, \[ \boxed{1} \]
Step 1: Since the centres are on the \(x\)-axis, let the centre be: \[ (a,0) \]
Step 2: Since the circle passes through the origin, its radius is: \[ r=a \]
Step 3: Equation of the circle is: \[ (x-a)^2+y^2=a^2 \] Expanding: \[ x^2-2ax+a^2+y^2=a^2 \] \[ x^2+y^2-2ax=0 \]
Step 4: This family contains only one arbitrary constant \(a\).
So we need to differentiate once to eliminate \(a\).
Step 5: Differentiate: \[ x^2+y^2-2ax=0 \] \[ 2x+2y\frac{dy}{dx}-2a=0 \] The highest derivative present is: \[ \frac{dy}{dx} \] So the order is: \[ 1 \] Therefore, \[ \boxed{1} \]
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