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One side of square is $$y=x+k$$ Distance of $$(5,3)$$ to the line $$y=x+k$$ is

$$\begin{aligned} & \frac{|3-5-k|}{\sqrt{2}}=\sqrt{2} \\ & =|-2-k|=2 \\ & \Rightarrow k=0 \text { or } k=-4 \end{aligned}$$

So lines are $$y=x$$ and $$y=x-4$$

Now, solving these lines with circle

$$\begin{aligned} y= & x \text { and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & 2 x^2-16 x+30=0 \\ \Rightarrow & x=3, y=3 \\ & x=5, y=5 \\ & y=x-4 \text { and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & x=5, y=1 \\ & x=7, y=3 \\ & \sum_{i=1}^4 x_i^2+y_i^2=9+9+25+25+25+1+49+9 \\ = & 152 \end{aligned}$$

19:T509,

JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Circle Question 34 Hindi Explanation

वर्ग की एक भुजा $$y=x+k$$ है बिंदु $$(5,3)$$ से रेखा $$y=x+k$$ की दूरी है

$$\begin{aligned} & \frac{|3-5-k|}{\sqrt{2}}=\sqrt{2} \\ & =|-2-k|=2 \\ & \Rightarrow k=0 \text { या } k=-4 \end{aligned}$$

अतः रेखाएँ $$y=x$$ और $$y=x-4$$ हैं

अब, इन रेखाओं को वृत्त के साथ हल करने पर

$$\begin{aligned} y= & x \text { और } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & 2 x^2-16 x+30=0 \\ \Rightarrow & x=3, y=3 \\ & x=5, y=5 \\ & y=x-4 \text { और } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & x=5,

Circle

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