Question

The parametric equation of a circle is given by \[x=3\cos \phi +2,\,y=3\sin \phi \]. Then its
\[\begin{align}
  & \text{A}\text{.Centre}=\left( -2,0 \right) \\
 & B.\text{Radius}=3 \\
 & C.\text{Centre}=\left( 2,0 \right) \\
 & D.\text{Radius}=1 \\
\end{align}\]

Answer 1

Hint: As $x,y$ are in terms of \[\phi \]. Try to find values of $\sin \phi ,\cos \phi $ in terms of \[x,y\]. Now substitute them into the very known trigonometric identity. By this you get an equation of circle. From where you can use the formula of coordinate geometry. If a circle equation is given by \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c\] center is \[\left( -g,-f \right)\] and radius is \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].

Complete step by step answer:
Given parametric form of the circle, can be written as:
\[\begin{align}
  & x=3\cos \phi +2\ldots \ldots \ldots ..\text{ }\left( \text{1} \right) \\
 & y=3\sin \phi \ldots \ldots \ldots \ldots \ldots \ldots \text{ }\left( \text{2} \right) \\
\end{align}\]
By taking equation (1) separately we can write it in the form:
\[x=3\cos \phi +2\]
By subtracting with 2 on both sides, we get it as of form,
\[x-2=3\cos \phi +2-2\]
By simplifying the above equation, we get it as:
\[x-2=3\cos \phi \]
By squaring on both sides of equation, we get it as:
\[{{\left( x-2 \right)}^{2}}={{\left( 3\cos \phi \right)}^{2}}\]
By simplifying the above equation, we get it as:
\[{{\left( x-2 \right)}^{2}}=9{{\cos }^{2}}\phi \ldots \ldots \ldots \ldots \ldots \text{ }\left( \text{3} \right)\]
By taking equation (2) separately, we get it in the form of:
\[y=3\sin \phi \]
By squaring on both sides of equation, we get it as:
\[{{y}^{2}}=9{{\sin }^{2}}\phi \ldots \ldots \ldots \ldots \ldots \left( \text{4} \right)\]
By basic general trigonometric knowledge, we can say the identity:
\[{{\cos }^{2}}\phi +{{\sin }^{2}}\phi =1\]
By multiplying with 9 on both sides of equation, we get it as:
\[9{{\cos }^{2}}\phi +9{{\sin }^{2}}\phi =9\]
By substituting equation (4) into above equation, we get it as:
\[{{\left( x-2 \right)}^{2}}+{{y}^{2}}=9\]
We know, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
By expanding first term and subtracting 9 on both sides, we get:
\[{{x}^{2}}+4-4x+{{y}^{2}}-9=0\]
By simplifying the above equation, we get it in form:
\[{{x}^{2}}+{{y}^{2}}-4x-5=0\]
We know for a circle with equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] center is \[\left( -g,-f \right)\] and radius is \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].
By comparing both the equations, we get value of g, f, c as:
\[g=\dfrac{-4}{2}=-2\,,\,\,\,f=0\,\,,\,\,\,c=-5\]
By substituting it above, we get center, radius as:
\[\text{Center}=\left( -\left( -2 \right),0 \right)\,\,;\,\,\,\text{Radius}\,=\sqrt{{{2}^{2}}-\left( -5 \right)}=\sqrt{4+5}\]
By simplifying we can say their values to be as:
\[\text{Center}\,=\,\left( 2,0 \right)\,\,;\,\,\text{Radius}\,=3\]

So, the correct answers are “Option B and C”.

Note: Be careful while taking the parametric form. The idea of leaving equation at stage of \[9{{\cos }^{2}}\phi \,,\,\,9{{\sin }^{2}}\phi \] is the idea to avoid fractions. It is very important that you stop there or else it will become long to solve. But try to minimize as much as possible.