TS EAMCET · Maths · Circle
The centre of the circle passing through the point \((0,1)\) and touching the curve \(y=x^2\) at \((2,4)\) is
- A \(\left(\frac{16}{5}, \frac{53}{10}\right)\)
- B \(\left(\frac{-2}{3}, \frac{-4}{3}\right)\)
- C \(\left(\frac{-4}{3}, \frac{2}{3}\right)\)
- D \(\left(\frac{-16}{5}, \frac{53}{10}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{-16}{5}, \frac{53}{10}\right)\)
Step-by-step Solution
Detailed explanation
Equation of tangent to the curve \(y=x^2\) at \((2,4)\) is \(\frac{y+4}{2}=2 x\) \[ \Rightarrow \quad 4 x-y-4=0 \] Equation of circle touching at \((2,4)\) and tangent \(4 x-y-4=0\) is \[ (x-2)^2+(y-4)^2+\lambda(4 x-y-4)=0 \] Since, this circle is also passes through \((0,1)\)…
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