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The real problem of mathematics in vocational high school
Let the circle equation be (x-a)? +(y-b)? =r?

The straight line tangent to the straight line 3x+4y-2=0 and vertical is y+ 1=(4/3)(x-2), that is, 4x-3y-1= 0.

The center (a, b) is on this straight line, and the distance to point P is d=r, that is:

①4a-3b- 1 1=0,②|3a-4b-2|/√(3? +4? )=r

Substitute y=0 into the cyclic equation and get: (x-a)? +b? =r? ,x 1=a+√(r? -B? ),x2=a-√(r? -B? )

The chord length obtained by cutting the positive semi-axis of X axis is 8=|x 1-x2|=2√(r? -B? ), which is r? -B? = 16 ③

① ② ③ Simultaneous solution: r=5, a=5, b=3.

So the circle equation is (x-5)? +(y-3)? =25

Go and see for yourself after work @ @ @