A square is inscribed in the circle x2+y2-10x-6y+30=0. One side of this square is parallel to y=x+3. If (xi, yi ) are the vertices of the square, then (xi2+yi2 ) is equal to:
- 152
- 148
- 156
- 160
Answer
(1) 152
Solution
Centre (5, 3), radius r = 25 + 9 - 30 = 2, so half the side of the square is r 2 = 2.
Since the sides are parallel or perpendicular to y = x + 3, take them as x - y + c = 0 and x + y + d = 0, each at distance 2 from the centre:
|5 - 3 + c| 2 = 2 c = 0, -4
|5 + 3 + d| 2 = 2 d = -6, -10
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So the sides are y = x, y = x - 4, x + y = 6, x + y = 10, and their meeting points give the vertices (3, 3), (5, 5), (5, 1), (7, 3).
(xi2 + yi2 ) = 18 + 50 + 26 + 58 = 152
Answer: (1)