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  • The parabola and the circle <img alt="\mathrm{(x-6)^{2}+y^{2}=r^{2}}" src="

The parabola \mathrm{y^{2}=4 x} and the circle \mathrm{(x-6)^{2}+y^{2}=r^{2}} will have no common tangent if \mathrm{'r'} is equal to

Option: 1

\mathrm{r>\sqrt{20}}


Option: 2

\mathrm{r<\sqrt{20}}


Option: 3

\mathrm{r>\sqrt{18}}


Option: 4

\mathrm{r \in(\sqrt{20}, \sqrt{28})}


Answers (1)

best_answer

Any normal of parabola is

\mathrm{y=-t x+2 t+t^{3}}
If it pass through \mathrm{(6,0)}, then


\mathrm{-6 t+2 t+t^3=0}
\mathrm{\Rightarrow \quad t=0, t^2=4, A \equiv(4,4)}.

Thus for no common tangent
\mathrm{ A C>\sqrt{4+16}>r }
\mathrm{ \Rightarrow \quad r>\sqrt{20} }

Posted by

Shailly goel

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