Question

The parametric equation of a circle is given by $x=3\cos \phi +2,\,y=3\sin \phi$. Then its

& \text{A}\text{.Centre}=\left( -2,0 \right) \\\ & B.\text{Radius}=3 \\\ & C.\text{Centre}=\left( 2,0 \right) \\\ & D.\text{Radius}=1 \\\ \end{aligned}$$

Answer 1

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:

& 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{aligned}$$ By taking equation (1) separately we can write it in the form: $$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.