Exam

JEE Main

Year

2019

Date

11 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Circle

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Circle

A square is inscribed in the circle x^ 2 +y^ 2 -6 x+8 y-103=0 with...

A square is inscribed in the circle $x^{2}+y^{2}-6 x+8 y-103=0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is:

Step-by-step solution

The equation of circle is, $x^{2}+y^{2}-6 x+8 y-103=0$ $\Rightarrow(x-3)^{2}+(y+4)^{2}=(8 \sqrt{2})^{2}$ $C(3,-4), r=8 \sqrt{2}$ $\Rightarrow$ Length of side of square $=\sqrt{2} r=16$
$\Rightarrow P(-5,4), Q(-5,-12)$ $R(11,-12), S(11,4)$ $\Rightarrow$ Required distance $=O P$ $=\sqrt{(-5-0)^{2}+(-4-0)^{2}}=\sqrt{25+16}=\sqrt{41}$