JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The radius of circle, having minimum area, which touches the cruve \(y = 4 - {x^2}\) and the lines \(y = \left| x \right|\) is :
- A \(4\left( {\sqrt 2 + 1} \right)\)
- B \(2\left( {\sqrt 2 + 1} \right)\)
- C \(2\left( {\sqrt 2 - 1} \right)\)
- D \( \frac{{ - 2 + \sqrt {34} }}{{2\sqrt 2 }}\)
Answer & Solution
Correct Answer
(D) \( \frac{{ - 2 + \sqrt {34} }}{{2\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
The equation has error (Let the equation of circle be \({x^2} + {\left( {y - k} \right)^2} = {r^2}\)) It touches \(x - y = 0\) \( \Rightarrow \left| {\frac{{0 - k}}{{\sqrt 2 }}} \right| = r\) \( \Rightarrow k = r\sqrt 2 \) \(\therefore \) Equation of circle bacomes…
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