A square is inscribed in the circle . One side of this square is parallel to y = x + 3. If are the vertices of the square, then is equal to [2024]

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Solution

Q.
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 [2024]

1 160

2 152

3 156

4 148

Ans.
(2)

One side of square is y = x + k

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

$\frac{| 3 - 5 - k |}{\sqrt{2}} = \sqrt{2}$

$\Rightarrow | - 2 - k | = 2 \Rightarrow k = 0 or k = - 4$

So lines are y = x and y = x – 4

Now, solving these lines with circle

y = x and $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$

$\Rightarrow 2 x^{2} - 16 x + 30 = 0$

$\Rightarrow$ x = 3, y = 3 and x = 5, y = 5

and y = x – 4 and $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$

$\Rightarrow$ x = 5, y = 1 and x = 7, y = 3

$\sum_{i = 1}^{4} x_{i}^{2} + y_{i}^{2}$ = 9 + 9 + 25 + 25 + 25 + 1 + 49 + 9 = 152.

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