The co-ordinates of the point equidistant from the positions of the three students are (-2.333, 0.667). The equation of the locus of the circular path is (x + 2.333)^2 + (y - 0.667)^2 = 31.889.

<p>Given the coordinates (-6, 5), (-3, -4), and (2, 1), we want to find the center and equation of the circle passing through these three points.</p><p>First, we need to remember that the center of a circle lies at the intersection of the perpendicular bisectors of any two chords of the circle. So, we need to find two chords that intersect and their perpendicular bisectors.</p><p>Let's choose the chords formed by points (-6, 5) and (-3, -4) and points (-6, 5) and (2, 1). To find the midpoint of each chord, we add the x-coordinates and divide by 2 to get the x-coordinate of the midpoint, and we add the y-coordinates and divide by 2 to get the y-coordinate of the midpoint.</p><p>Midpoint of chord 1:</p><p>(((-6) + (-3)) / 2, ((5) + (-4)) / 2) = (-4.5, 0.5)</p><p>Midpoint of chord 2:</p><p>(((-6) + (2)) / 2, ((5) + (1)) / 2) = (-2, 3)</p><p>Now, we need to find the slope of the line passing through each chord. The slope of a line passing through two points (x1, y1) and (x2, y2) is (y2 - y1) / (x2 - x1).</p><p>Slope of chord 1:</p><p>((-4) - (5)) / ((-3) - (-6)) = -3/2</p><p>Slope of chord 2:</p><p>((1) - (5)) / ((2) - (-6)) = -4/8 = -1/2</p><p>The slope of the perpendicular bisector of each chord is the negative reciprocal of the slope of the chord. So, the slope of the perpendicular bisector of chord 1 is 2/3 (the negative reciprocal of -3/2) and the slope of the perpendicular bisector of chord 2 is -2 (the negative reciprocal of -1/2).</p><p>To find the equation of the perpendicular bisector of each chord, we can use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the midpoint of the chord and m is the slope of the perpendicular bisector.</p><p>Equation of the perpendicular bisector of chord 1:</p><p>y - 0.5 = (2/3)(x + 4.5)</p><p>Equation of the perpendicular bisector of chord 2:</p><p>y - 3 = -2(x + 2)</p><p>Now, we can solve these two equations simultaneously to find the coordinates of the center of the circle. We can do this by setting the right-hand sides of the two equations equal to each other, since they both equal y.</p><p>(2/3)(x + 4.5) + 0.5 = -2(x + 2) + 3</p><p>Simplifying this equation, we get:</p><p>2x + 3y + 13 = 0</p><p>This is the equation of the perpendicular bisector of both chords, so it must pass through the center of the circle. Solving this equation for y, we get:</p><p>y = (-2/3)x - (13/3)</p><p>So, the center of the circle is (x, y) = (1, -3).</p>

<p>To find the coordinates of the point equidistant from the positions of the three students, we can find the circumcenter of the triangle formed by their positions. The circumcenter is the point equidistant from all three vertices of the triangle.</p><p><br></p><p>First, let's label the coordinates of the students as follows:</p><p>Student 1: (-6, 5)</p><p>Student 2: (-3, -4)</p><p>Student 3: (2, 1)</p><p><br></p><p>To find the coordinates of the circumcenter, we can use the following steps:</p><p><br></p><p>Find the midpoints of two sides of the triangle:</p><p><br></p><p>Midpoint between Student 1 and Student 2: ((-6 + (-3))/2, (5 + (-4))/2) = (-4.5, 0.5)</p><p>Midpoint between Student 2 and Student 3: ((-3 + 2)/2, (-4 + 1)/2) = (-0.5, -1.5)</p><p>Find the slopes of the lines passing through the sides of the triangle:</p><p><br></p><p>Slope of the line passing through Student 1 and Student 2: (5 - (-4))/(-6 - (-3)) = 9/-3 = -3</p><p>Slope of the line passing through Student 2 and Student 3: (-4 - 1)/(-3 - 2) = -5/-5 = 1</p><p>Find the perpendicular bisectors of the sides:</p><p><br></p><p>Perpendicular bisector of the line passing through Student 1 and Student 2:</p><p><br></p><p>Slope of the perpendicular bisector = -1/slope of the line passing through Student 1 and Student 2 = 1/3</p><p>Using the midpoint (-4.5, 0.5) and the slope 1/3, we can write the equation of the perpendicular bisector as:</p><p>y - 0.5 = (1/3)(x + 4.5)</p><p>Simplifying the equation: 3y - 1.5 = x + 4.5</p><p>Rearranging the equation: x - 3y + 6 = 0</p><p>Perpendicular bisector of the line passing through Student 2 and Student 3:</p><p><br></p><p>Slope of the perpendicular bisector = -1/slope of the line passing through Student 2 and Student 3 = -1</p><p>Using the midpoint (-0.5, -1.5) and the slope -1, we can write the equation of the perpendicular bisector as:</p><p>y + 1.5 = -1(x + 0.5)</p><p>Simplifying the equation: y + 1.5 = -x - 0.5</p><p>Rearranging the equation: x + y + 2 = 0</p><p>Solve the simultaneous equations formed by the two perpendicular bisectors:</p><p><br></p><p>x - 3y + 6 = 0</p><p>x + y + 2 = 0</p><p>Solving these equations, we find the coordinates of the circumcenter: (x, y) = (-2, 4)</p><p><br></p><p>Therefore, the coordinates of the point equidistant from the positions of the three students are (-2, 4).</p><p><br></p><p>To find the equation of the locus of the circular path, we need to find the equation of the circle passing through the three given points. The equation of a circle can be represented as (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius.</p><p><br></p><p><br></p><p>Using the coordinates of the circumcenter (-2, 4) as the center of the circle, we need to find the radius.</p><p><br></p><p>To find the radius, we can use the distance formula between the circumcenter (-2, 4) and any of the student's positions. Let's choose Student 1 (-6, 5) as a reference:</p><p><br></p><p>Distance between circumcenter and Student 1:</p><p>r = sqrt((-2 - (-6))^2 + (4 - 5)^2)</p><p>= sqrt(4^2 + (-1)^2)</p><p>= sqrt(16 + 1)</p><p>= sqrt(17)</p><p><br></p><p>Therefore, the radius of the circle is sqrt(17).</p><p><br></p><p>The equation of the locus of the circular path is:</p><p>(x - (-2))^2 + (y - 4)^2 = (sqrt(17))^2</p><p>(x + 2)^2 + (y - 4)^2 = 17</p><p>(x + 2)^2 + (y - 4)^2 = 17</p><p><br></p><p>So, the equation of the locus of the circular path is (x + 2)^2 + (y - 4)^2 = 17.</p>