(x-2)^2+(y-2)^2=2
Then AB is the tangent of circle p and the tangent point is e.
According to the circular power theorem
ae^2=ap^2-r^2=(a-2)^2+2^2-2^2=(a-2)^2
be^2=bp^2-r^2=(b-2)^2+2^2-2^2=(b-2)^2
AB^2=a^2+b^2
And AB=AE+BE.
So a 2+b 2 = (a-2+b-2) 2.
2ab-8(a+b)+ 16=0
(a-4)*(b-4)= a B- 4(a+b)+ 16 = 8
Let the m coordinate of the midpoint of AB be (x, y).
So x=a/2 and y=b/2.
We can get (2x-4)(2y-4)=8 from (1).
So the trajectory equation of m is (x-2)(y-2)=2.
The estimation is to find the area of ABO, and O is the origin.
S△ABO=ab/2
It can be obtained from 2ab-8(a+b)+ 16=0.
ab/2=2(a+b)-4≥4√ab-4
ab-8√ab+8≥0
(√ab-4)^2≥8
√ab-4≥2√2
ab≥24+ 16√2
Therefore, S△ABO≥ 12+8√2.
(In fact, when finding the extreme value, it is a=b=4+2√2. You can go in and check it out. )
In fact, this problem can be transformed into an inscribed circle in which the circle (x-2) 2+(y-2) 2 = 2 is △AOB, and the locus of the midpoint of AB can be found. But in contrast, the original problem is simpler.
the second question
If m≠0
Kl 1=m, Kl2=- 1/m, so l1⊥ L2;
L 1 always crosses the origin o,
L2 can be changed to x-2y+(m+2)(y- 1)=0, that is, L2 always passes through the intersection a of x-2y=0 and y- 1=0, and the coordinates of the intersection a of x-2y=0 and y- 1=0 are (2
If m=0
L 1 is y=0, L2 is x=2, and the coordinate of intersection p is (2,0), which is also in the above trajectory.
To sum up, the trajectory equation of P is (x-1) 2+(y-1/2) 2 = 5/4.
The other intersection of L 1 and circle p, P 1, is the origin, and the other intersection of L2 and circle p, P2, is a, which is (2, 1).
Obviously, when PB⊥P 1P2, S△PP 1P2 is the largest, and PB=P 1P2/2.
Therefore, the maximum value of S△PP 1P2 = Pb * p 1p2/2 = ao 2/4 = 5/4.
P 1P2 indicates that the linear equation of AO is x-2y=0, so the slope of the straight line PB is -2, and the linear equation is y- 1/2=-2(x- 1).
Combine the solutions of y- 1/2=-2(x- 1) and (x-1) 2+(y-1/2) 2 = 5/4, and you get it.
x- 1= 1/2
Therefore, x=3/2, y=- 1/2, m=y/x=- 1/3.
Or x= 1/2, y=3/2 and m=3.
Generally speaking, it is very troublesome to solve these two problems by hard calculation. Although the words are wordy, the calculation is small and clear.