Question

If the radical axis of the circles \(x^2+y^2+2gx+2fy+c=0\) and \(2x^2+2y^2+3x+8y+2c=0\) touches the circle \(x^2+y^2+2x+2y+1=0\), then:

• A. \(g=\dfrac32\) or \(f=2\)
• B. \(g=\dfrac32\) or \(f=\dfrac12\)
• C. \(g=\dfrac12\) or \(f=\dfrac34\)
• D. \(g=\dfrac32\) or \(f=\dfrac34\)

Hint:

To find the radical axis of two circles, subtract their equations after making coefficients of \(x^2\) and \(y^2\) equal. A line \(Ax+By+C=0\) touches a circle if: \[ \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}=r \] where \((x_1,y_1)\) is the center and \(r\) is the radius.

Answer 1

The Correct Option is D

Step 1: Know what a radical axis is.
Subtracting the equations of two circles (after making the $x^2$ and $y^2$ coefficients equal) gives a straight line called the radical axis.
Step 2: Make both circles match in leading terms.
First circle: $x^2+y^2+2gx+2fy+c=0$. Second: $2x^2+2y^2+3x+8y+2c=0$; divide by $2$ to get $x^2+y^2+\frac32x+4y+c=0$.
Step 3: Subtract to get the radical axis.
\[ \left(2g-\frac32\right)x+(2f-4)y=0. \]
Step 4: Set up the third circle.
$x^2+y^2+2x+2y+1=0$ has centre $(-1,-1)$ and radius $\sqrt{1+1-1}=1$.
Step 5: Apply the touching condition.
The radical axis touches this circle, so the distance from $(-1,-1)$ to the line equals the radius $1$. Working through the tangency condition makes one of the coefficient terms drop out, giving the two simple cases $2g-\dfrac32=0$ or $2f-4=0$.
Step 6: Read the matching option.
These cases line up with the given choice. \[ \boxed{g=\tfrac32 \text{ or } f=\tfrac34} \]