Question

Show that the polar equation \(r = a\sin \theta + b\cos \theta \)_{, where}\(ab \ne 0\)_{, represents a circle, and find its center and radius.}

Short answer

Centre of the circle is \(\left( {\frac{b}{2},\frac{a}{2}} \right)\) and radius is \(r = \frac{{\sqrt {{b^2} + {a^2}} }}{2}\) .

Step-by-step solution

Step 1:

We have \(r = a\sin \theta + b\cos \theta \) -----(1)

Since\(x = r\cos \theta \)and\(y = r\sin \theta \)

Then \(\frac{x}{r} = \cos \theta \) and \(\frac{y}{r} = \sin \theta \)

Step 2:

Putting the values of \(\cos \theta \) and \(\sin \theta \) in equation(1)

We have\(r = \frac{{ay}}{r} + \frac{{bx}}{r}\)

\( \Rightarrow {r^2} = ay + bx\)

Since\({r^2} = {x^2} + {y^2}\), then

\( \Rightarrow {x^2} - bx + {y^2} - ay = 0\)

\( \Rightarrow {x^2} - bx + \frac{{{b^2}}}{4} - \frac{{{b^2}}}{4} + {y^2} - ay + \frac{{{a^2}}}{4} - \frac{{{a^2}}}{4} = 0\)(making perfect squares)

\( \Rightarrow {\left( {x - \frac{b}{2}} \right)^2} + {\left( {y - \frac{a}{2}} \right)^2} - \frac{{{b^2}}}{4} - \frac{{{a^2}}}{4} = 0\)

\( \Rightarrow {\left( {x - \frac{b}{2}} \right)^2} + {\left( {y - \frac{a}{2}} \right)^2} = \frac{{{b^2} + {a^2}}}{4}\)

\({\left( {x - \frac{b}{2}} \right)^2} + {\left( {y - \frac{a}{2}} \right)^2} = \frac{{{b^2} + {a^2}}}{4}\) Represents a circle

Step 3:

Comparing with \({\left( {x - A} \right)^2} + {\left( {y - B} \right)^2} = {r^2}\)

Centre of the circle is \(\left( {\frac{b}{2},\frac{a}{2}} \right)\) and radius is \(r = \frac{{\sqrt {{b^2} + {a^2}} }}{2}\) .