JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{C}\) be the circle of minimum area touching the parabola \(y=6-x^2\) and the lines \(y=\sqrt{3}|x|\). Then, which one of the following points lies on the circle \(C\)?
- A \((2,4)\)
- B \((1,2)\)
- C \((2,2)\)
- D \((1,1)\)
Answer & Solution
Correct Answer
(A) \((2,4)\)
Step-by-step Solution
Detailed explanation
Equation of circle \(x^2+(y-(6-r))^2=r^2\) touches \(\sqrt{3} \mathrm{x}-\mathrm{y}=0\) \( p=r \) \( \frac{|0-(6-r)|}{2}=r \) \( |r-6|=2 r \) \( r=2\) \(\therefore\) Circle \(\mathrm{x}^2+(\mathrm{y}-4)^2=4\) \((2,4)\) Satisfies this equation
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