A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ One side of the square is parallel to $y = x + 3$ Then determine the co-ordinates of its vertices.

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Solution

Circle
is (5, 3), 2 AB being parallel to y = x + 3 makes 45 $^{\circ}$ with x-axis so that AC will make 45 $^{\circ}$ + 45 $^{\circ}$ =90 $^{\circ}$ with x-axis
$∴$ $\dfrac {x \, - \, 5 }{cos ^{\circ}}$
$= \frac{y - 3}{s i n 90^{o}}$

$= 2, - 2 f o r C, A$
$∴$ $n x = 5, y = 3 + 2 o r 3 - 2$
$∴$ (5 , 5 ) and (5 , 1) are two vertices C and A . DB will make an angle of $0^{\circ}$ with x - axis
$∴$ $\frac{x - 5}{c o s 0^{\circ}} = \frac{x - 3}{s i n 0^{\circ}} = 2, - 2$ for B , D

$∴$ $x = 5 + 2, 5 - 2, y = 3$
$∴$ $(7, 3), (3, 3)$ are the vertices B and D

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A square is inscribed in the circle x2+y2−10x−6y+30=0 One side of the square is parallel to y=x+3 Then determine the co-ordinates of its vertices.

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