Answer

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Hint: First eliminate the trigonometric values, i.e., eliminate $\theta $. Then find the general equation using other terms. And compare with the general equations of the straight line or the circle.

Complete step-by-step answer:

Now let’s consider the following equation given in the question;

$\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$

Now let’s consider,

$\dfrac{x-{{x}_{1}}}{\cos \theta }=r$

So,

$x-{{x}_{1}}=r\cos \theta $ …………..(i)

Now let’s consider,

$\dfrac{y-{{y}_{1}}}{\sin \theta }=r$

So,

$y-{{y}_{1}}=r\sin \theta $ …………..(ii)

Now let consider the equations (i) and (ii) and square and add them;

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta $

Taking out the common term, we get

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)$

We know the trigonometric identity, $\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right)$, substituting this value in the above equation, we get

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}........(ii)$

We know the general equation of circle is ${{(x-a)}^{2}}+{{(x-b)}^{2}}={{r}^{2}}$ , here (a, b) are the centre of the circle and ‘r’ is the radius of the circle.

Comparing equation (ii) with the general equation of circle, we get

$({{x}_{1}},{{y}_{1}})$ as the centre and $'r'$ as the radius of the circle.

Hence, $\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$ represents equation of circle with a known centre and a given radius

So, option (d) is the correct answer.

Note: In this type of equation, there is always a confusion of solving parametric equations, so one should try to eliminate terms in terms of $'\theta '$.

Another approach is squaring the given ratios directly and then adding them, this will also give the same approach.

Common mistake made by students is when they see the equation ${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta $, they start expanding the left hand side. This will be a tedious and lengthy process.

Complete step-by-step answer:

Now let’s consider the following equation given in the question;

$\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$

Now let’s consider,

$\dfrac{x-{{x}_{1}}}{\cos \theta }=r$

So,

$x-{{x}_{1}}=r\cos \theta $ …………..(i)

Now let’s consider,

$\dfrac{y-{{y}_{1}}}{\sin \theta }=r$

So,

$y-{{y}_{1}}=r\sin \theta $ …………..(ii)

Now let consider the equations (i) and (ii) and square and add them;

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta $

Taking out the common term, we get

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)$

We know the trigonometric identity, $\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right)$, substituting this value in the above equation, we get

${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}........(ii)$

We know the general equation of circle is ${{(x-a)}^{2}}+{{(x-b)}^{2}}={{r}^{2}}$ , here (a, b) are the centre of the circle and ‘r’ is the radius of the circle.

Comparing equation (ii) with the general equation of circle, we get

$({{x}_{1}},{{y}_{1}})$ as the centre and $'r'$ as the radius of the circle.

Hence, $\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$ represents equation of circle with a known centre and a given radius

So, option (d) is the correct answer.

Note: In this type of equation, there is always a confusion of solving parametric equations, so one should try to eliminate terms in terms of $'\theta '$.

Another approach is squaring the given ratios directly and then adding them, this will also give the same approach.

Common mistake made by students is when they see the equation ${{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta $, they start expanding the left hand side. This will be a tedious and lengthy process.

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