It means that the slope of the straight line passing through point P is 1.

Satisfy y y Y = x+k (k < 0).

x-y+k=0

The radius of R is 4, and the distance to the straight line is also 4.

Let a straight line intersect with point F.

Then according to the formula of the distance from the point to the straight line, R=|4-4+k|/√ 1+ 1=4.

K= soil 4√2, ∴k=-4√2.

The linear equation is y=x-4√2.

Ac = bc ab is the diameter, then ∠ BAC = 45 pay attention to RT⊥AC.

Then r=RT=√2/2AR=2√2.

Let T(x, y)

satisfy

{|x-y-4√2|/√2=2√2

{√ (x-4) 2+(y-4) 2 = (2 √ 2) (distance from t to r)

{x=

{y=

Solve the equation and find x, y,

Let PM be y=kx+b again

According to the common tangent and the distance between the two centers and the straight line is equal to the radius, two equations are listed and can be solved. When two common tangents intersect at a point, point P is found. The process is a bit cumbersome. After playing so much, I hope to adopt O(∩_∩)O!