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 (x i , y i ) are the vertices of the square, then sum (x i 2 +y i 2 ) is equal to :

Mathematics · Circles

JEE Main 2024 — 4 April, Shift 1 — Question 17

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 $\sum\left(x_{i}^{2}+y_{i}^{2}\right)$ is equal to :

$y=x+c \quad \& \quad x+y+d=0$

$\left|\frac{5-3+c}{\sqrt{2}}\right|=\sqrt{2}$

$\quad\left|\frac{8+\mathrm{d}}{\sqrt{2}}\right|=\sqrt{2}$

$|c+2|=2$

$8+\mathrm{d}= \pm 2$

$\mathrm{c}=0,-4$

$d=-10,-6$

pts $(5,5),(3,3),(7,3),(5,1)$

$\sum\left(x_{i}^{2}+y_{1}^{2}\right)=25+25+9+9+49+9+25+1$ $=152$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.