Point \( (2,1) \) and \( (6,7) \) are opposite vertices of a square which is nscribed in a circle find the centre and equation of a circle

To find the center of the circle inscribing the square, we first need to find the midpoint of the segment connecting the opposite vertices \( (2,1) \) and \( (6,7) \). The midpoint \( M \) is calculated as follows: \[ M = \left( \frac{2+6}{2}, \frac{1+7}{2} \right) = \left( \frac{8}{2}, \frac{8}{2} \right) = (4, 4) \] The center of the circle is at the point \( (4, 4) \). Next, we find the distance between the two opposite vertices to determine the radius. The distance is calculated using the distance formula: \[ d = \sqrt{(6-2)^2 + (7-1)^2} = \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \] The distance represents the diagonal of the square, and the radius of the circle is half of the diagonal: \[ \text{Radius} = \frac{d}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13} \] Finally, the equation of the circle is given by \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h,k) \) is the center and \( r \) is the radius. Thus, we have: \[ (x - 4)^2 + (y - 4)^2 = 13 \] This is the equation of the circle inscribing the square. To sum it up, the center of the circle is \( (4, 4) \) and the equation of the circle is \( (x - 4)^2 + (y - 4)^2 = 13 \).