Question

Find the centre of the circle where \[x = 2 + 3\cos \theta \] and \[y = 3\sin \theta - 1\]
A. \[\left( {3,3} \right)\]
B. \[\left( {2, - 1} \right)\]
C. \[\left( { - 2,1} \right)\]
D. \[\left( {1, - 2} \right)\]

Answer 1

Hint: We will be transforming the equations using trigonometric identity in such a way that we are able to see both the equation in the form of the general equation of a circle.

Formula Used: We will be using the general form of a circle which is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] where \[\left( {h,k} \right)\] is the center and \[r\] is a radius of circle.
Sum formula of trigonometry: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]

Complete step by step solution:
Let us assume that,
\[x = 2 + 3\cos \theta \] ------ (i)
And
\[y = 3\sin \theta - 1\] ----- (ii)
Simplifying equation (i) and we get
\[x = 2 + 3\cos \theta \]
\[x - 2 = 3\cos \theta \] ------ (iii)
Simplifying equation (ii) and we get
\[y = 3\sin \theta - 1\]
\[y + 1 = 3\sin \theta \] ----- (iv)
Squaring both equation (iii) and (iv) in order to get the equations as general form of circle.
From equation (iii), we get
\[x - 2 = 3\cos \theta \]
\[{\left( {x - 2} \right)^2} = 9{\cos ^2}\theta \] ----- (v)
From equation (iv), we get
\[y + 1 = 3\sin \theta \]
\[{\left( {y + 1} \right)^2} = 9{\sin ^2}\theta \] ----- (vi)
Now adding both the squared equations (v) and (vi) and we get ,
\[{\left( {x - 2} \right)^2} + {\left( {y + 1} \right)^2} = 9{\cos ^2}\theta + 9{\sin ^2}\theta \]
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y + 1} \right)^2} = 9\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\]
We know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\],
Substitute \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] and we get
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - ( - 1)} \right)^2} = 9\] -- (vii)
Equating the equation (vii) with the general form \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] and we will get the center in the form of \[\left( {h,k} \right)\].
\[\therefore \left( {2, - 1} \right)\] is the centre of circle.

Hence the answer is B which is\[\left( {2, - 1} \right)\].

Note: While solving the question student must remember the basic standard form of a circle to equate the equation and get the coordinates of the centre through the equation. Also need to take care while we square the trigonometric functions. If there have any constant function then this also should be squared. e.g., if we have an expression \[a\sin x\] then square of this expression should be \[{\left( {a\sin x} \right)^2} = {a^2}{\sin ^2}x\] .