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If the equation x2 + Y2 + 4mx-2y + 4m2 + 4m = 0 represents a circle, then the value range of real number m is
The solution is x2 + Y2 + 4mx-2y + 4m2 + 4m = 0
We get x2 + 4mx + 4m2 + y2-2y + 4m = 0
That is, (x-2m) ^ 2 + (Y-1) ^ 2 = 1-4m
The circle is represented by the equation x2 + Y2 + 4mx-2y + 4m2 + 4m = 0
Then 1-4m ＞ 0
M < 1 / 4
The equation of a straight line passing through M (2,1) and tangent to the circle x2 + y2-6x-8y + 24 = 0 is
(x-3)²+(y-4)²=1
Center (3,4), radius 1
The distance from the center of the circle to the tangent is equal to the radius
X = 2
The slope exists
y-1=k(x-2)
kx-y+1-2k=0
|3k-4+1-2k|/√(k²+1)=1
k²-6k+9=k²+1
k=4/3
So X-2 = 0 and 4x-3y-5 = 0
The equation of a straight line passing through point m (2,1) and tangent to circle x ^ 2 + y ^ 2-6x-8y + 24 = 0 is
The answer is 4x-3y-5 = 0, x = 2 for the complete problem-solving process
x^2+y^2-6x-8y+24=0
(x-3) ^ 2 + (y-4) ^ 2 = 1, center of circle is (3,4), radius is 1
Let the linear equation be Y-1 = K (X-2), kx-y + 1-2k = 0
The distance from the center of the circle to the straight line is
|3K - 4 + 1-2k | / radical (k ^ 2 + 1) = | K-3 | / radical (k ^ 2 + 1) = radius = 1
k = 4/3
4X - 3Y - 5 = 0 or x = 2 (where k is infinite)
A linear equation which is tangent to all circles represented by equation (1 + λ) x2 + (1 + λ) Y2 + 6x-8y-25 (λ + 3) = 0
A linear equation which is tangent to all circles represented by equation (1 + λ) x & # 178; + (1 + λ) y & # 178; + 6x-8y-25 (λ + 3) = 0
(1+λ)[x²+6x/(1+λ)]+(1+λ)[y²-8/(1+λ)]-25(λ+3)=0
(1+λ)[x+3/(1+λ)]²+(1+λ)[y-4/(1+λ)]²-25/(1+λ)-25(λ+3)=0
[x+3/(1+λ)]²+[y-4/(1+λ)]²-25/(1+λ)²-25(λ+3)/(1+λ)=0
[x+3/(1+λ)]²+[y-4/(1+λ)]²=[5(λ+2)/(1+λ)]²
Center (- 3 / (1 + λ), 4 / (1 + λ)), radius r = 5 (λ + 2) / (1 + λ)
Since [4 / (1 + λ) / [- 3 / (1 + λ)] = - 4 / 3 = constant, the centers of all circles are on the line y = - (4 / 3) x passing through the origin
The distance from the center of the circle to the origin d = 5 / (1 + λ), R-D = 5 (λ + 2) / (1 + λ) - 5 / (1 + λ) = (5 λ + 5) / (1 + λ) = 5 = constant,
Therefore, all circles are inscribed at points (3, - 4), and the equation of the common tangent to all circles is y = (3 / 4) (x-3) - 4 = (3 / 4) x-25 / 4
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