Question

The radius of the circle $r = a\;\cos \theta + $ $b\sin \theta $ is
A). $\sqrt {\dfrac{{{a^2} + {b^2}}}{4}} $ B) \[\sqrt {{a^2} + {b^2}} \] C) $\dfrac{{{a^2} + {b^2}}}{2}$ D) ${a^2} + {b^2}$

Answer 1

Hint: We will find the standard form of the given equation and then we will compare it with the standard form of the equation of the circle and hence it will give the radius of the circle.

Complete step-by-step answer:
The general form of the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ whose centre is $\left( { - g, - f} \right)$ and radius is $\sqrt {{g^2} + {f^2} - c} $ where g, f, c are 3 constant.
Here, we are given $r = a\cos \theta + b\sin \theta = 0$ (1)
So, our approach will be to convert the given equation in some standard form and then comparing it, we can find the radius.
To solve, this problem, lets first put the value of $cos\theta = \dfrac{x}{r}$ and $\sin \theta = \dfrac{y}{r}$ for the easy approach to our solution.
$\therefore $ we know the trigonometric identity,
${\cos ^2}\theta + {\sin ^2}\theta = 1.$
Also, we know, $\sin \theta = {\raise0.5ex\hbox{$\scriptstyle p$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle h$}}$
          \[\cos \theta = {\raise0.5ex\hbox{$\scriptstyle b$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle h$}}\]
So, putting the values,
          ${\sin ^2}\theta + {\cos ^2}\theta = 1$
          $ \Rightarrow {\left( {\dfrac{p}{h}} \right)^2} + {\left( {\dfrac{b}{h}} \right)^2} = 1$
          $ \Rightarrow \dfrac{{{p^2} + {b^2}}}{{{h^2}}} = 1$

$ \Rightarrow \dfrac{{{h^2}}}{{{h^2}}} = 1$
     1 = 1
     Hence, LHS = RHS.
     We get $2gx = - ax$
     \[ \Rightarrow g = - {\raise0.5ex\hbox{$\scriptstyle a$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}};\]
     $2fy = - by$
     $ \Rightarrow 2f = - b$
     $f = - {\raise0.5ex\hbox{$\scriptstyle b$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}},$
     C = 0
Radius \[ = \sqrt {{g^2} + {f^2} - c} \]
     $ = \sqrt {{{\left( { - {\raise0.5ex\hbox{$\scriptstyle a$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}^2} + {{\left( {\dfrac{{ - b}}{2}} \right)}^2} - 0} $
     \[ = \sqrt {\dfrac{{{a^2}}}{4} + \dfrac{{{b^2}}}{4}} \]
     \[ = \sqrt {\dfrac{{{a^2} + {b^2}}}{4}} \]
     $ = \sqrt {\dfrac{{{a^2} + {b^2}}}{4}} $ (A).

Note: The equation of the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ represents the radius that is equal to \[\sqrt {{g^2} + {f^2} - c} \]