Exam

JEE Main

Year

2024

Date

4 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Circle

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Circle

A square is inscribed in the circle x^2+y^2-10 x-6 y+30=0 . One...

A square is inscribed in the circle $x^2+y^2-10 x-6 y+30=0$. One side of this square is parallel to $y=x+3$. If \left(x_i, y_i\right)$are the vertices of the square, then$\mathbf{\Sigma}\left(x_i^2+y_i^2\right) is equal to:

Step-by-step solution

$y=x+c \quad \&$ $$\begin{array}{ll} y=x+c \quad & & x+y+d=0 \\ \left|\frac{5-3+c}{\sqrt{2}}\right|=\sqrt{2} & \left|\frac{8+d}{\sqrt{2}}\right|=\sqrt{2} \\ |c+2|=2 & 8+d= \pm 2 \\ c=0,-4 & d=-10,-6 \end{array}$$ pts $(5,5),(3,3),(7,3),(5,1)$ $$\begin{aligned} & \sum\left(\mathrm{x}_{\mathrm{i}}^2+\mathrm{y}_1^2\right)=25+25+9+9+49+9+25+1 \\ & =152 \end{aligned}$$