If point P(4, 2) lies on the circle having centre O(-1, 1), and C is the circumference of the circle, then value of \(49C^{2}\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 50336
P lies on the circle, so the radius is the distance OP, found by the distance formula. The circumference is then \(C=2\pi r\), and \(C^{2}=4\pi^{2}r^{2}\) needs only \(r^{2}\), so no square root is needed.
Square of the radius (distance formula): \(r^{2}=(4+1)^{2}+(2-1)^{2}=26\)
Square of the circumference: \(C^{2}=4\pi^{2}\times26=104\pi^{2}\)
Putting in \(\pi^{2}=\frac{484}{49}\): \(\displaystyle 49C^{2}=104\times484=50336\)
The factor 49 is there to cancel the \(7^{2}\) in \(\pi^{2}=\frac{484}{49}\), which is why the question takes \(\pi=\frac{22}{7}\).