According to the vertical knowledge, k*k 1=- 1.
The slope of the chord is k=-2,
And through the chord passing through point A, the linear equation can be obtained by the point oblique method: y-(-2)=-2(x-4).
Simplification: 2x+y-6=0
2. Solution: Do it according to the geometric relationship.
The standard equation of a circle is (x- 1)? +(y- 1)？ = 1, so the center o coordinate is (1, 1) and the radius is 1.
The distance from the center O to the straight line L is d = L3×1+4×1+8l/√ (3? +4? )= 3
According to the geometric meaning diagram, we can see that:
The minimum distance is d radius 1=2.
The maximum distance is d+ radius 1=4.
3. Solution: If the slope of the straight line L does not exist, due to the passing point (-5,-10).
So the linear equation is x=-5.
At this time, the straight line is tangent to the circle and has no chord length. So the slope of the straight line exists.
Let the slope of the straight line L be k and the equation be y-(- 10)=kx-(-5).
That is kx-y+5k- 10=0.
According to the equation of a circle, we know that the center of the circle is (0,0) and the radius is 5.
According to the geometric relationship, the distance from the center O to the straight line L is:
D= 5 under the root sign? -(5√2/2)=5√2/2
The distance formula from a point to a straight line, the distance from O to L is
d=5√2/2=l5k- 10l/√(k？ + 1)
The solution is k= 1 or k=7.
So the equation of the straight line L is x-y-5=0 or 7x-y+25=0.
4. Solution: Because the chord length of the tangent line I of the circle C is 2√3, according to the geometric relationship.
The distance from the center of the circle to the chord under the root sign is d= 2? -(√3)? = 1
According to the equation of circle C, the center C: (a, 2) and the radius r=2 are known.
According to the distance from the center of the circle C to the straight line L: X-Y+3 = 0 is 1, the distance formula from point to straight line is applied.
d= 1=la-2+3l/√2
The solution is a=- 1+√2 or a=- 1-√2.