Question

The locus of the center of a circle of radius 3 which rolls on the outside of a circle \[{x^2} + {y^2} + 3x - 6y - 9 = 0\] is
A) \[{x^2} + {y^2} + 3x - 6y - 45 = 0\]
B) \[{x^2} + {y^2} + 3x - 6y - 31 = 0\]
C) \[{x^2} + {y^2} + 3x - 6y + 5 = 0\]
D) \[{x^2} + {y^2} + 3x - 6y + \dfrac{{29}}{4} = 0\]

Answer 1

Hint:
We will first consider the given equation of a circle that is \[{x^2} + {y^2} + 3x - 6y - 9 = 0\]. Next, we will make the complete square of the given equation and convert it into the form \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]. As we are given that we have to find the locus of the Centre of a circle of radius 3 so, we will add 3 in the obtained radius to find the locus of the Centre outside the circle. After this, we will simplify the equation by opening the squares. Hence, we will get the desired result.

Complete step by step solution:
We will first consider the equation given in the question that is \[{x^2} + {y^2} + 3x - 6y - 9 = 0\].
As we have to find the locus of the Centre of the circle, we will first find the equation of the circle of the form \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] by completing the squares on the right hand side of the equation.
Thus, we will first complete the square of \[x\] by adding \[{\left( {\dfrac{3}{2}} \right)^2}\] on both the sides,
\[
   \Rightarrow {x^2} + 3x + {\left( {\dfrac{3}{2}} \right)^2} + {y^2} - 6y = 9 + {\left( {\dfrac{3}{2}} \right)^2} \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {y^2} - 6y = \dfrac{{45}}{4} \\
 \]
Next, we will complete the square of \[y\] by adding \[{\left( 3 \right)^2}\] on both the sides of the above-obtained equation, we get,
\[
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {y^2} - 6y + {\left( 3 \right)^2} = \dfrac{{45}}{4} + {\left( 3 \right)^2} \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {\left( {y - 3} \right)^2} = \dfrac{{45}}{4} + 9 \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {\left( {y - 3} \right)^2} = \dfrac{{81}}{4} \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {\left( {y - 3} \right)^2} = {\left( {\dfrac{9}{2}} \right)^2} \\
 \]
Further, we have to find the locus of the centre of the circle of radius 3, so, we will add 3 on the right-hand side of the obtained equation which is the radius part and as we have to find the locus of the circle of radius 3 so, we have added it in the obtained radius \[\dfrac{9}{2}\].
Hence, we get,
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + {\left( {y - 3} \right)^2} = {\left( {\dfrac{9}{2} + 3} \right)^2}\]
Now, we will simplify the above equation by opening the square.
\[
   \Rightarrow {x^2} + \dfrac{9}{4} + 3x + {y^2} + 9 - 6y = \dfrac{{225}}{4} \\
   \Rightarrow {x^2} + {y^2} + 3x - 6y - 45 = 0 \\
 \]
Thus, we get the required locus equation.

Hence, (A) is the correct option.

Note:
Complete the square roots properly by adding the same number on both the sides of the equation. While simplifying the brackets, make sure you have opened all the squares properly. As we have to find the locus of the circle of radius 3, so we have to add the given radius in the given equation of the circle. Remember the general form of equation of circle that is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\].