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Question: If \(\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r\), then the equation repr...

If x−x1cos⁡θ=y−y1sin⁡θ=r\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r, then the equation represents:
A. Equation of a straight line, if θ\theta is constant and rr is variable
B. Equation of a circle, if rr is constant and θ\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

Explanation

Solution

Hint: First eliminate the trigonometric values, i.e., eliminate θ\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;
x−x1cos⁡θ=y−y1sin⁡θ=r\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r
Now let’s consider,
x−x1cos⁡θ=r\dfrac{x-{{x}_{1}}}{\cos \theta }=r
So,
x−x1=rcos⁡θx-{{x}_{1}}=r\cos \theta …………..(i)
Now let’s consider,
y−y1sin⁡θ=r\dfrac{y-{{y}_{1}}}{\sin \theta }=r
So,
y−y1=rsin⁡θy-{{y}_{1}}=r\sin \theta …………..(ii)
Now let consider the equations (i) and (ii) and square and add them;
(x−x1)2+(y−y1)2=r2cos⁡2θ+r2sin⁡2θ{{(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−x1)2+(y−y1)2=r2(cos⁡2θ+sin⁡2θ){{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)
We know the trigonometric identity, (cos⁡2θ+sin⁡2θ=1)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right), substituting this value in the above equation, we get
(x−x1)2+(y−y1)2=r2........(ii){{(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=r2{{(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
(x1,y1)({{x}_{1}},{{y}_{1}}) as the centre and ′r′'r' as the radius of the circle.
Hence, x−x1cos⁡θ=y−y1sin⁡θ=r\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 ′θ′'\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−x1)2+(y−y1)2=r2cos⁡2θ+r2sin⁡2θ{{(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.