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

Question

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 $$\Sigma\left(x_i^2+y_i^2\right)$$ is equal to:

Accepted Answer

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

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

Solutions · 1