Question

The locus of the centre of a circle of radius 2 which rolls on the outside of the circle \( x^{2}+y^{2}+3x-6y-9=0 \) along its circumference is

• A. \( x^{2}+y^{2}+3x-6y+5=0 \)
• B. \( x^{2}+y^{2}+3x-6y-31=0 \)
• C. \( x^{2}+y^{2}+3x-6y+\frac{29}{4}=0 \)
• D. \( x^{2}+y^{2}-3x+6y+31=0 \)

Hint:

The equation of any concentric circle differs from the original circle only by the constant term \( c \). You can skip expanding the whole equation by using the shortcut: \( c_{\text{new}} = c_{\text{old}} + r_1^2 - R^2 \).

Answer 1

The Correct Option is B

Step 1: Picture what is happening.
A small circle of radius 2 rolls along the outside edge of a fixed circle. As it rolls, its centre always stays the same distance away from the centre of the fixed circle. So the path (locus) traced by the small circle's centre is itself a circle that has the same centre as the fixed circle.
Step 2: Find the centre of the fixed circle.
The fixed circle is $x^{2}+y^{2}+3x-6y-9=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$, we read $2g=3$ and $2f=-6$, so $g=\frac{3}{2}$ and $f=-3$. The centre is $(-g,-f)$. \[ \text{Centre} = \left(-\tfrac{3}{2},\ 3\right) \]
Step 3: Find the radius of the fixed circle.
The radius is $\sqrt{g^2+f^2-c}$ with $c=-9$. \[ r_1=\sqrt{\tfrac{9}{4}+9+9}=\sqrt{\tfrac{81}{4}}=\tfrac{9}{2} \]
Step 4: Find the radius of the locus circle.
Since the small circle (radius 2) rolls on the outside, the gap between the two centres equals the two radii added together. \[ R=r_1+2=\tfrac{9}{2}+2=\tfrac{13}{2} \]
Step 5: Write the locus circle.
It shares the centre $\left(-\tfrac{3}{2},3\right)$ and has radius $\tfrac{13}{2}$. \[ \left(x+\tfrac{3}{2}\right)^2+(y-3)^2=\left(\tfrac{13}{2}\right)^2 \]
Step 6: Expand and simplify.
Opening the squares gives $x^2+3x+\tfrac{9}{4}+y^2-6y+9=\tfrac{169}{4}$. Move everything to one side: the constant becomes $\tfrac{9}{4}+9-\tfrac{169}{4}=-31$. \[ \boxed{x^{2}+y^{2}+3x-6y-31=0} \]