Two diagonals of rectangle ABCD intersect at M (2,0) AB side, the linear equation is x-3y-6 = 0 point t (- 1,1) on the straight line of AD, the moving circle P passes through n Two diagonals of rectangle ABCD intersect at the edge of M (2,0) AB, and the linear equation is x-3y-6 = 0 point t (- 1,1). On the straight line where ad is located, the moving circle P passes through n (- 2,0) and is circumscribed with the circumscribed circle of ABCD, and the trajectory equation of the center of the moving circle is obtained

Two diagonals of rectangle ABCD intersect at M (2,0) AB side, the linear equation is x-3y-6 = 0 point t (- 1,1) on the straight line of AD, the moving circle P passes through n Two diagonals of rectangle ABCD intersect at the edge of M (2,0) AB, and the linear equation is x-3y-6 = 0 point t (- 1,1). On the straight line where ad is located, the moving circle P passes through n (- 2,0) and is circumscribed with the circumscribed circle of ABCD, and the trajectory equation of the center of the moving circle is obtained


Because the moving circle P passes through the point n, so | PN | is the radius of the circle, and because the moving circle P is the outer circle of the circle m, so | PM | = | PN | + 2, that is | PM | - | PN | = 2; therefore, the trajectory of point P is a circle with m, n as the focus and the real axis length of 2. Because the real half axis length a = root sign 2, the half focal length C = 2. The imaginary half axis length b = root sign 2



It is known that the center of the square is m (1,4) and a vertex is a (0,2)


The linear equation of Ma is y = 2x + 2 and the slope is 2
Then the equation of the line passing through M and perpendicular to Ma is
Y = - 0.5x + 4.5 (the other two intersections of the square are on this line)
Make a circle with m point as the center and Ma as the radius
(X-1)^2+(Y-4)^=(4-2)^2+(1-0)^2=5
Intersection with y = - 0.5x + 4.5
Two intersections are the other two vertices of the square
B(3,3)C(-1,5)
The AB equation is y = x / 3 + 2
The AC equation is y = - 3x + 2



Make a straight line through the point m (0,1) so that the line segment cut by two straight lines L1: x-3y + 10 = 0, L2: 2x + Y-8 = 0 is exactly bisected by m, and solve the linear equation


Let B (T, 8-2t) and m (0, 1) be the midpoint of AB, and a (- t, 2t-6) can be obtained from the midpoint coordinate formula



Make a straight line through the point m (0,1) so that the line segment cut by two straight lines L1: x-3y + 10 = 0, L2: 2x + Y-8 = 0 is exactly bisected by m, and solve the linear equation


Let the obtained line and the known lines L1 and L2 intersect at two points a and B respectively. ∵ point B is on the line l2:2x + Y-8 = 0, so let B (T, 8-2t). And m (0, 1) is the midpoint of AB, and a (- t, 2t-6) is obtained from the midpoint coordinate formula. ∵ point a is on the line L1: x-3y + 10 = 0, ∵ (- t) - 3 (2t-6) + 10 = 0, and T = 4. ∵ B (4, 0), a (- 4, 2), so the obtained linear equation is: x + 4y-4 =0.