Given the circle equation x ^ 2 + y ^ 2-2ax-4ay + 5A ^ 2-4 = 0 (a is not equal to 0), we can determine the linear equation that no matter what the value of a is cut by the circle, the chord length is 1

Given the circle equation x ^ 2 + y ^ 2-2ax-4ay + 5A ^ 2-4 = 0 (a is not equal to 0), we can determine the linear equation that no matter what the value of a is cut by the circle, the chord length is 1

x^2+y^2-2ax-4ay+5a^2-4=0
(x-a)^2+(y-2a)^2=4
So, as a changes, the center of the circle moves on y = 2x
Therefore, the straight line with chord length of 1 cut by circle is parallel to y = 2x, which is set as y = 2x + B
If the chord length is 1, the distance from the center of the circle to the chord = √ (4 - (1 / 2) ^ 2 = √ 15 / 2
(0,0) is one of the centers
So, | B | / √ (1 + 4) = √ 15 / 2
b=±5√3/2
Therefore, the linear equation is y = 2x ± 5 √ 3 / 2

Given circle C: x ^ 2 + y ^ 2-2ax-4ay + 9A ^ 2 / 2 = 0 (a > 0), find the total tangent line

x²+y²-2ax-4ay+9a²/2=0
x²-2ax+a²+y²-4ay+4a²=a²/2
(x-a)²+(y-2a)²=a²/2
Center coordinates (a, 2a), a > 0, radius = A / √ 2
The straight line perpendicular to the x-axis and its tangent: x = a + √ (A & # 178 / 2), x = a - √ (A & # 178 / 2), which is related to the value of a, and is rounded off
In other cases, let the linear equation y = KX + B
kx-y+b=0
The line is tangent to the circle, the distance from the center of the circle to the line = radius
|ka-2a+b|/√(1+k²)=a/√2
It's time to tidy up
a²(1+k²)=2[(k-2)a+b]²
(k²-8k+7)a²+4b(k-2)a+2b²=0
For any a > 0, the equation holds only if
k²-8k+7=0 (1)
4b(k-2)=0 (2)
2b²=0 (3)
From (1), we get (k-1) (k-7) = 0
K = 1 or K = 7
From (3), B = 0
K = 1, B = 0; k = 7, B = 0, substituting (2), the equation holds
To sum up, we get two linear equations y = x, y = 7x

It is known that the center of circle C is on the x-axis and passes through point a (- 4,3) if the line L: 2ax-y + 8A + 1 = 0
It is known that the center of the circle C is on the x-axis and passes through the point a (- 4,3) if the line L: 2ax-y + 8A + 1 = 0. Proof: no matter what real number a takes, the line l always intersects the circle C

I haven't played this kind of math problem for ten years, but it doesn't seem difficult. First, let the straight line L: x = - 4, you can get y = 1; then the straight line l must pass through the point (- 4,1); then prove that the point (- 4,1) is in the circle C

It is proved that no matter what real number m takes, the line l always intersects the circle C
It is known that the line L: 2mx-y-8m-3 = 0 and the circle C: (x-3) ^ 2 + (y + 6) ^ 2 = 25
It is proved that no matter what real number m takes, the line l always intersects the circle C
The equation for finding the shortest line L of line segment cut by circle C
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2mx-y-8m-3=0
2m(x-4)-y-3=0
Over a (4, - 3)
(4-3)^2+(-3+6)^2=10