KCET · Maths · Circle
If \( x=2+3 \cos \theta \) and \( y=1-3 \sin \theta \) represent a circle then the centre and radius is
- A \( (2,1), 9 \)
- B \( (2,1), 3 \)
- C \( (1,2), \frac{1}{3} \)
- D \( (-2,-1), 3 \)
Answer & Solution
Correct Answer
(B) \( (2,1), 3 \)
Step-by-step Solution
Detailed explanation
Given that,
\(x=2+3 \cos \theta \rightarrow(1)\)
\(y=1-3 \sin \theta \rightarrow(2)\)
From Eqs. \((1)\) and \((2)\), we have
\(\frac{x-2}{3}=\cos \theta\) and \(\frac{y-1}{-3}=\sin \theta\)
We know that, \(\sin ^{2} \theta+\cos ^{2} \theta=1\)
So, \(\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-1}{-3}\right)=1\)
\(\Rightarrow \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{9}=1\)
\(\Rightarrow(x-2)^{2}+(y-1)^{2}=9\)
So, it is circle centre at point \((2,1)\) and radius is 3 .
\(x=2+3 \cos \theta \rightarrow(1)\)
\(y=1-3 \sin \theta \rightarrow(2)\)
From Eqs. \((1)\) and \((2)\), we have
\(\frac{x-2}{3}=\cos \theta\) and \(\frac{y-1}{-3}=\sin \theta\)
We know that, \(\sin ^{2} \theta+\cos ^{2} \theta=1\)
So, \(\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-1}{-3}\right)=1\)
\(\Rightarrow \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{9}=1\)
\(\Rightarrow(x-2)^{2}+(y-1)^{2}=9\)
So, it is circle centre at point \((2,1)\) and radius is 3 .
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