Question

If the equation \[ ax^2+by^2+2hxy+2gx+2fy+c=0 \] represents a circle passing through the origin, then

• A. \(a=b,\ c=0\)
• B. \(|a|=|b|,\ h=0=c\)
• C. \(a=b,\ h=c=0\)
• D. \(a=b,\ h=0\)

Hint:

For a second degree equation to represent a circle: \[ a=b \] and \[ h=0. \] If the circle passes through the origin, additionally: \[ c=0. \]

Answer 1

The Correct Option is C

Step 1: Recall the general equation of a circle.
The general equation of a circle is \[ x^2+y^2+2gx+2fy+c=0 \] Comparing with \[ ax^2+by^2+2hxy+2gx+2fy+c=0, \] for a circle we must have \[ a=b \] and \[ h=0 \]

Step 2: Use the condition that the circle passes through the origin.
Since the circle passes through \[ (0,0), \] substitute \[ x=0,\quad y=0 \] in the equation: \[ a(0)^2+b(0)^2+2h(0)(0)+2g(0)+2f(0)+c=0 \] \[ c=0 \]

Step 3: Combine all conditions.
Thus, the required conditions are \[ a=b,\quad h=0,\quad c=0 \]

Step 4: Match with the options.
This corresponds to option \[ (3) \]

Step 5: Why \(a=b\)?
A circle has equal coefficients of \[ x^2 \] and \[ y^2 \] because the radius is same in all directions.

Step 6: Why \(h=0\)?
The term \[ xy \] appears in rotated conics, but a circle has no \(xy\)-term in standard orientation.

Step 7: Final conclusion.
Therefore, \[ \boxed{a=b,\ h=c=0} \]