Question

Find the center of the circle $x = - 1 + 2\cos \theta $ , $y = 3 + 2\sin \theta $
A. $\left( {1, - 3} \right)$
B. $\left( { - 1,3} \right)$
C. $\left( {1,3} \right)$
D. None of these

Answer 1

Hint: First we will find the value of $\cos \theta $ and $\sin \theta $ from the given equation. Then take their square and add them.

Formula Used:
We will be using the standard equation of the circle which is ${(x - h)^2}\; + {\left( {y - k} \right)^2}\; = {{{r}}^2}$ where, $\left( {h,k} \right)$ are the center of the circle.

Complete step by step solution:
We will be transforming both equations in such a way that it should be in the form of the standard form of a circle.
Now,
$x = - 1 + 2\cos \theta $
$ \Rightarrow {\left( {x + 1} \right)^2} = {\left( {2\cos \theta } \right)^2}$
$ \Rightarrow {\left( {x + 1} \right)^2} = 4{\cos ^2}\theta $ ------ (i)
Now we will be transforming the equation of $y$ in the same format-
$y = 3 + 2\sin \theta $
$ \Rightarrow {\left( {y - 3} \right)^2} = 4{\sin ^2}\theta $ ----- (ii)
Adding equations (i) and (ii)
$ \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} = 4\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)$
$ \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} = 4$
We have found the final equation by adding equations (i) and (ii),
Now we will be equating the calculated equation with the standard form of a circle to find out the center of the circle.
$ \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} = 4$
${(x - h)^2}\; + {\left( {y - k} \right)^2}\; = {{{r}}^2}$
From this $\left( {h,k} \right) = \left( { - 1,3} \right)$

Option ‘B’ is correct

Note: Always find the coordinates (h, k) of the center of the circle and radius ‘r’ by converting the given equation to the standard form of general equation which will easily solve the question.