MHT CET · Maths · Circle
The centre and radius of a circle \(x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}\), are respectively
- A \((0,0)\) and \(3 a\) units
- B \((0,0)\) and \(4 a\) units
- C \((0,0)\) and \(2 a\) units
- D \((0,0)\) and \(a\) units
Answer & Solution
Correct Answer
(B) \((0,0)\) and \(4 a\) units
Step-by-step Solution
Detailed explanation
Given equation of circle in parametric form is
\(
x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right) \text { and } y=4 a\left(\frac{2 t}{1+t^{2}}\right)
\)
\(x=4 a \cos 2 \theta\) and \(y=4 a \sin 2 \theta\), where \(t=\tan \theta\)
Comparing with \(x=r \cos \theta, y=r \sin \theta\), we get \(r=4 a\), centre is \((0,0)\)
\(
x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right) \text { and } y=4 a\left(\frac{2 t}{1+t^{2}}\right)
\)
\(x=4 a \cos 2 \theta\) and \(y=4 a \sin 2 \theta\), where \(t=\tan \theta\)
Comparing with \(x=r \cos \theta, y=r \sin \theta\), we get \(r=4 a\), centre is \((0,0)\)
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