Question

The equations of the circle which pass through the origin and makes intercepts of lengths \(4\) and \(8\) on the \(x\)-axis and \(y\)-axis respectively are

• A. \(x^2+y^2\pm 4x\pm 8y=0\)
• B. \(x^2+y^2\pm 2x\pm 4y=0\)
• C. \(x^2+y^2\pm 8x\pm 16y=0\)
• D. \(x^2+y^2\pm x\pm y=0\)

Hint:

For circle through origin: \(x^2+y^2+2gx+2fy=0\). Intercepts on axes are \(|2g|\) and \(|2f|\).

Answer 1

The Correct Option is A

Step 1: General equation of circle through origin.
\[ x^2+y^2+2gx+2fy=0 \]
Step 2: Find intercepts.
On \(x\)-axis, put \(y=0\):
\[ x^2+2gx=0 \Rightarrow x(x+2g)=0 \]
Intercept length on \(x\)-axis is \(|2g|\). Given \(=4\).
So:
\[ |2g|=4 \Rightarrow g=\pm 2 \]
On \(y\)-axis, put \(x=0\):
\[ y^2+2fy=0 \Rightarrow y(y+2f)=0 \]
Intercept length on \(y\)-axis is \(|2f|\). Given \(=8\).
So:
\[ |2f|=8 \Rightarrow f=\pm 4 \]
Step 3: Substitute values.
\[ x^2+y^2+2(\pm 2)x+2(\pm 4)y=0 \]
\[ x^2+y^2\pm 4x\pm 8y=0 \]
Final Answer:
\[ \boxed{x^2+y^2\pm 4x\pm 8y=0} \]