Center C(- 1, 2), radius r=√2.
① The tangent line passes through the coordinate origin, and the tangent line AB:kx-y=0.
The distance d from the center of the circle C(- 1, 2) to the tangent AB:kx-y=0 is equal to the radius r,
∴d=|-k-2|/√( 1+k^)=√2.
∴k=2-√6 or k=2+√6.
Tangent equation: y=(2-√6)x or y = (2+√ 6) x.
② Let A(a, 0), B(0, a), a≠0.
Tangent AB:x/a+y/a= 1, that is, x+y-a=0.
d=|- 1+2-a|/√2=√2。
|a- 1|=2。
A=- 1 or a=3.
Tangent equation: x+y+ 1=0 or x+y-3=0.
So there are four tangent equations: y=(2-√6)x or y=(2+√6)x or x+y+ 1=0 or x+y-3=0.
(2) as shown in figure | pm | =| PC |-cm | = | po |
[(x 1+ 1)^+(y 1-2)^]-2=x 1^+y 1^
2x 1-4y 1+3=0。 Point p satisfies the equation: 2x-4y+3=0.
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∴ Line PO Vertical line 2x 1-4y 1+3=0. That is, the vertical line of the straight line PC 2x 1-4y 1+3=0.
Linear PC:y=-2x instead of 2x 1-4y 1+3=0.
x 1=-3/ 10，y 1=3/5
∴P(-3/ 10,3/5)