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  • The equation of a circle is . The centre of the smallest circl

The equation of a circle is \mathrm{ x^2+y^2=4}. The centre of the smallest circle touching this circle and the line \mathrm{ x+y=5 \sqrt{2}} has the coordinates

Option: 1

\left(\frac{7}{2 \sqrt{2}}, \frac{7}{2 \sqrt{2}}\right)


Option: 2

\left(\frac{3}{2}, \frac{3}{2}\right)


Option: 3

\left(-\frac{7}{2 \sqrt{2}},-\frac{7}{2}\right)


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Here, \mathrm{\mathrm{OB}= radius =2}

The distance of  \mathrm{(0,0)} from \mathrm{x+y=5 \sqrt{2}} is 5 .

\mathrm{\therefore}  the radius of the smallest circle \mathrm{=\frac{5-2}{2}=\frac{3}{2}}

and \mathrm{\mathrm{OC}=2+\frac{3}{2}=\frac{7}{2}}

The slope of \mathrm{\mathrm{OA}=1=\tan \theta}

\mathrm{ \begin{aligned} & \therefore \cos \theta=\frac{1}{\sqrt{2}}, \sin \theta=\frac{1}{\sqrt{2}} \\ & \therefore C=(0+\text { OC } \cdot \cos \theta, 0+\text { OC } \cdot \sin \theta) \end{aligned} }

=\left(\frac{7}{2 \sqrt{2}}, \frac{7}{2 \sqrt{2}}\right)

Posted by

Rakesh

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