JEE Main 2024 (Online) 4th April Morning ShiftQuestion 1
A square is inscribed in the circle $x^{2}+y^{2}-10x-6y+30=0$. One side of this square is parallel to $y=x+3$. If $(x_i,y_i)$ are the vertices of the square, then $\sum\left(x_i^{2}+y_i^{2}\right)$ is equal to :
Circle
Solution
Concept. Write each vertex as centre $+$ a vector $v_i$ of length $r$; for a square $\sum v_i=0$. Formulas used. (1) $|c+v|^2=|c|^2+2c\cdot v+|v|^2$Step 1 - Circle. Centre $c=(5,3)$, $r^2=25+9-30=4$, $r=2$. Step 2 - Use (1). Vertices $c+v_i$, $|v_i|=2$, and opposite vertices give $v_3=-v_1$, $v_4=-v_2$, so $\sum v_i=0$: $$\sum|c+v_i|^2=4|c|^2+2c\cdot\textstyle\sum v_i+\sum|v_i|^2=4\cdot34+0+16=152.$$ Step 3 - Check. A side parallel to $y=x$ means diagonals parallel to the axes: vertices $(7,3),(3,3),(5,5),(5,1)$. Sum $=58+18+50+26=152$. Answer: B