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If $\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$, then the equation represents:
A. Equation of a straight line, if $\theta $ is constant and $r$ is variable
B. Equation of a circle, if $r$ is constant and $\theta $ is a variable
C. A straight line passing through a fixed point and having a known slope
D. A circle with a known centre and a given radius

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Last updated date: 15th Jul 2024
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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.