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Cannot use the formula from point to line! ! ! A very hateful math problem! ! !
L:y=(3/4)x+8,3x-4y+32=0,k=3/4

y=0,x=-32/3。 One (-32/3,0)

x=0,y=8。 B(0,8)

(1) Find the coordinates of point p and the radius r of ⊙ p.

PB:kPB=- 1/k=-4/3

y=-4x/3+8

y=0,x=6,P(6,0)

Ob = 8, op = 6, and the circle centered on p is tangent to the straight line L at point B.

R=PB= 10

(2) If ⊙P moves to the left along the X axis at 3/ 10 units per second, and the radius of ⊙P decreases at 2/3 units per second, let ⊙P move for t seconds, and ⊙P always intersects with the straight line L, try to find the range of t;

R≥ the distance from point P to line L, then ⊙P always intersects with line L. 。

p,R= 10-2t/3,L:3x-4y+32=0

Distance from point P to line L H=| 10-9t/50|

10-2t/3≥|

10-2t/3≥ 10-9t/50≥-( 10-2t/3)

t≤0

There is something wrong with the topic. Try to change 3/ 10 to 10/3.

Distance from point P to line L: H=| 10-2t|

10-2t/3≥ 10-2t≥-( 10-2t/3)

7.5≥t≥0

(3) In (2), let ⊙P be cut by a straight line L with a chord length of A, and ask whether there is a value of t to maximize A? If it exists, find the value of t.

To maximize a, t must have a value.

(a/2)^2=r^2-h^2=( 10-2t/3)^2-( 10-2t)^2=(-32/9)*(t- 15/4)^2+50

T = 15/4,(a/2) 2 Max =50,a Max = 10√2。

(4) In (2), let the intersection of ⊙P and the straight line L be q, so that △APQ is similar to △ABO. Please write the value of t directly at this time.

△APQ is similar to△ △ABO, and PQ is perpendicular to AB.

⊙P is tangent to the straight line L.

T=0, or t=7.5