Question

Show that \[x=a\cos \alpha +b\sin \alpha ,y=a\sin \alpha -b\cos \alpha \] represents a circle for all values of \[\alpha \].

Answer 1

Hint: At first try to remove or eliminate ‘ \[\alpha \] '. Then try to find a relation or an equation between x and y.

Complete step-by-step answer:

In the question we are given the parametric forms of x and y as $\left( a\cos \alpha +b\sin \alpha \right)$ and $\left( a\sin \alpha -b\cos \alpha \right)$ respectively.

So at first we have to eliminate terms related to $\alpha $ from x and y to get an equation or a relation in terms of a and b only.

So let’s consider,

$x=a\cos \alpha +b\sin \alpha $ ………………………..(i)

$y=a\sin \alpha -b\cos \alpha $ ………………………(ii)

Now squaring equation (i) we get,

${{x}^{2}}={{\left( a\cos \alpha +b\sin \alpha \right)}^{2}}$

Here we will use the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

So, ${{x}^{2}}={{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha +2ab\sin \alpha \cos

\alpha $ ………………………………………(iii)

Now, squaring equation (ii) we get,

${{y}^{2}}={{\left( a\sin \alpha -b\cos \alpha \right)}^{2}}$

Here we will use the formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

So, ${{y}^{2}}={{a}^{2}}{{\sin }^{2}}\alpha +{{b}^{2}}{{\cos }^{2}}\alpha -2ab\sin \alpha \cos

\alpha $ ……………………………………..(iv)

Now as we know values of ${{x}^{2}}$ and ${{y}^{2}}$ we will add them up.

In the adding we will use the identity

${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$

So, by adding equation (iii) and (iv) we get

${{x}^{2}}+{{y}^{2}}={{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha +2ab\sin \alpha

\cos \alpha +{{a}^{2}}{{\sin }^{2}}\alpha +{{b}^{2}}{{\cos }^{2}}\alpha -2ab\sin \alpha \cos

\alpha $

Which can be written as,

${{x}^{2}}+{{y}^{2}}={{a}^{2}}\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha

\right)+{{b}^{2}}\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)$

${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$

We get the final equation as ${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$ ………………………..(v)

Let see the general equation of circle which is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

By comparing the equation (v) with general equation of circle we can conclude that

${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$ is also a equation of circle.

The equation of ${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$ is in form of circle.


Note: In this type of question there is always a confusion to solve. So, the 1st step is to eliminate the variable term like ‘ \[\alpha \] '.