JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(\sum\left(\mathrm{x}_{\mathrm{i}}^2+\mathrm{y}_{\mathrm{i}}^2\right)\) is equal to :
- A \(148\)
- B \(156\)
- C \(160\)
- D \(152\)
Answer & Solution
Correct Answer
(D) \(152\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}y=x+c \quad \& & x+y+d=0 \\ \left|\frac{5-3+c}{\sqrt{2}}\right|=\sqrt{2} & \left|\frac{8+d}{\sqrt{2}}\right|=\sqrt{2}\end{array}\) \(|c+2|=2 \)\( 8+d= \pm 2 \) \(c=0,-4 \)\( d=-10,-6 \) \(\operatorname{pts}(5,5),(3,3),(7,3),(5,1) \)…
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