If a circle (x-1)^2+y^2=r^2 touches the ellip | vedclass

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(A) Given circle is $(x-1)^2+y^2=r^2 \dots (i)$Here,the centre of the circle is $(1,0)$ and the radius is $r$.Given ellipse is $x^2+4y^2=16$,which can

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$\sqrt{\frac{11}{3}}$

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(A) Given circle is $(x-1)^2+y^2=r^2 \dots (i)$Here,the centre of the circle is $(1,0)$ and the radius is $r$.Given ellipse is $x^2+4y^2=16$,which can be written as $\frac{x^2}{16}+\frac{y^2}{4}=1$.Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=4$ and $b=2$.Since the circle touches the ellipse internally,the normal to the ellipse at the point of contact $P$ must pass through the centre of the circle $(1,0)$.Let the point of contact be $P(4\cos	heta, 2\sin	heta)$.The equation of the normal to the ellipse at $P$ is given by $ax\sec	heta - by\operatorname{cosec}	heta = a^2-b^2$.Substituting $a=4, b=2$ and $a^2-b^2 = 16-4=12$,we get $4x\sec	heta - 2y\operatorname{cosec}	heta = 12$.Since this normal passes through $(1,0)$,we have $4(1)\sec	heta - 2(0)\operatorname{cosec}	heta = 12$,which implies $4\sec	heta = 12$,so $\sec	heta = 3$.Thus,$\cos	heta = \frac{1}{3}$ and $\sin	heta = \sqrt{1-\left(\frac{1}{3}ight)^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.The point $P$ is $\left(4\left(\frac{1}{3}ight), 2\left(\frac{2\sqrt{2}}{3}ight)ight) = \left(\frac{4}{3}, \frac{4\sqrt{2}}{3}ight)$.The radius $r$ is the distance between the centre $(1,0)$ and the point $P\left(\frac{4}{3}, \frac{4\sqrt{2}}{3}ight)$.$r^2 = \left(\frac{4}{3}-1ight)^2 + \left(\frac{4\sqrt{2}}{3}-0ight)^2 = \left(\frac{1}{3}ight)^2 + \frac{32}{9} = \frac{1}{9} + \frac{32}{9} = \frac{33}{9} = \frac{11}{3}$.Therefore,$r = \sqrt{\frac{11}{3}}$.

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$

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