JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
The centre of the circle passing through the point \((0,1)\) and touching the parabola \(y=x^{2}\) at the point \((2,4)\) is
- A \(\left(\frac{3}{10}, \frac{16}{5}\right)\)
- B \(\left(\frac{-16}{5}, \frac{53}{10}\right)\)
- C \(\left(\frac{6}{5}, \frac{53}{10}\right)\)
- D \(\left(\frac{-53}{10}, \frac{16}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{-16}{5}, \frac{53}{10}\right)\)
Step-by-step Solution
Detailed explanation
\(y=x^{2}\) \(\left.\frac{d y}{d x}\right|_{P}=4\) \((y-4)=4(x-2)\) \(4 x-y-4=0\) Circle \(:(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0\) passes through (0,1) \(4+9+\lambda(-5)=0 \Rightarrow \lambda=\frac{13}{5}\) Circle \(: x^{2}+y^{2}+x(4 \lambda-4)+y(-\lambda-8)+(20-4 \lambda)=0\)…
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