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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 :

  1. A \(148\)
  2. B \(156\)
  3. C \(160\)
  4. D \(152\)
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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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