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Question: The radius of the circle \(r = a\;\cos \theta + \) \(b\sin \theta \) is A). \(\sqrt {\dfrac{{{a^2}...

The radius of the circle r=a  cos⁡θ+r = a\;\cos \theta + bsin⁡θb\sin \theta is
A). a2+b24\sqrt {\dfrac{{{a^2} + {b^2}}}{4}} B) a2+b2\sqrt {{a^2} + {b^2}} C) a2+b22\dfrac{{{a^2} + {b^2}}}{2} D) a2+b2{a^2} + {b^2}

Explanation

Solution

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 x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 whose centre is (−g,−f)\left( { - g, - f} \right) and radius is g2+f2−c\sqrt {{g^2} + {f^2} - c} where g, f, c are 3 constant.
Here, we are given r=acos⁡θ+bsin⁡θ=0r = 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 cosθ=xrcos\theta = \dfrac{x}{r} and sin⁡θ=yr\sin \theta = \dfrac{y}{r} for the easy approach to our solution.
∴\therefore we know the trigonometric identity,
cos⁡2θ+sin⁡2θ=1.{\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⁡θ=\raise0.5exb/\lower0.25exh\cos \theta = {\raise0.5ex\hbox{$\scriptstyle b$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle h$}}
So, putting the values,
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
⇒(ph)2+(bh)2=1\Rightarrow {\left( {\dfrac{p}{h}} \right)^2} + {\left( {\dfrac{b}{h}} \right)^2} = 1
⇒p2+b2h2=1\Rightarrow \dfrac{{{p^2} + {b^2}}}{{{h^2}}} = 1

⇒h2h2=1 \Rightarrow \dfrac{{{h^2}}}{{{h^2}}} = 1
1 = 1
Hence, LHS = RHS.
We get 2gx=−ax2gx = - ax
⇒g=−\raise0.5exa/\lower0.25ex2;\Rightarrow g = - {\raise0.5ex\hbox{$\scriptstyle a$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}};
2fy=−by2fy = - by
⇒2f=−b\Rightarrow 2f = - b
f = - {\raise0.5ex\hbox{\scriptstyle b} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}},
C = 0
Radius =g2+f2−c= \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}
=a24+b24= \sqrt {\dfrac{{{a^2}}}{4} + \dfrac{{{b^2}}}{4}}
=a2+b24= \sqrt {\dfrac{{{a^2} + {b^2}}}{4}}
=a2+b24= \sqrt {\dfrac{{{a^2} + {b^2}}}{4}} (A).

Note: The equation of the circle x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 represents the radius that is equal to g2+f2−c\sqrt {{g^2} + {f^2} - c}