K(OC)=(2-0)/(2- 1)=2

The equation is: y-0=2(x- 1).

That is, y=2x-2.

2) when the chord AB is bisected by the point p,

The line between the center of the circle C and the point P must be perpendicular to AB.

So we get the slope of AB.

k=- 1/2

y-2=- 1/2(x-2)

x+2y-6=0

3) The inclination of the straight line L is 45, and the equation of the straight line AB is y = x..

The distance from the center of the circle (1, 0) to the straight line y=x is1√ 2.

By using the vertical diameter theorem, it is obtained that |AB|=2×√34/2=√34.

2. 1) 1. Simultaneous linear equations 3x+4y-2=0 and L2: 2x+y+2 = 0.

Get p (-2,2)

Because the straight line passes through the origin

So y=-x

2)L3:X-2Y- 1=0 Slope k3=2.

So the slope of the linear equation is-1/2.

So the linear equation is

x+2=- 1/2(y-2)

2x+y+2=0

3.

Let M(X, Y)A(m, n) because m is the midpoint of AB, so we can get the relationship between m, n and x, y, m+3/2=x n+4/2=y, then we can get m=2x-3 n=2y-4, and then we can substitute m and n into the equation of the circle. +4(y-2)？ =4