The equation of circle C is known as x ^ 2 + y ^ 2-6c-8y + 21 = 0 The equation of straight line is kx-y - (4k-3) = 0 (1) It is proved that no matter why K is a real number, the line L and C must intersect (2) Let l be at the intersection of circle C, AB, what is k, and what is the minimum value of AB

The equation of circle C is known as x ^ 2 + y ^ 2-6c-8y + 21 = 0 The equation of straight line is kx-y - (4k-3) = 0 (1) It is proved that no matter why K is a real number, the line L and C must intersect (2) Let l be at the intersection of circle C, AB, what is k, and what is the minimum value of AB


Is there something wrong with your title? C in the circular equation should be y? If so, you can formulate the circular equation. You can see that the center point of the circle is a (3,4) and the radius is 2. Then you bring B (4,3) into the linear equation, and you will find that (4,3) holds no matter what the value of K is, that is, the line must pass through the point (4



The position relationship between circle O1: x2 + Y2 + 6x-7 = 0 and circle O2: x2 + Y2 + 6y-27 = 0 is______ .


Circle O1: x2 + Y2 + 6x-7 = 0, the standard equation is (x + 3) 2 + y2 = 16, the center is (- 3, 0), the radius is 4, circle O2: x2 + Y2 + 6y-27 = 0, the standard equation is x2 + (y + 3) 2 = 36, the center is (0, - 3), the radius is 6, the center distance is 32 ∵ 6-4 ∵ 32 ∵ 6 + 4, the two circles intersect



The equation x2 + Y2 - (4m + 2) x-2my + 4m2 + 5m + 2 = 0 represents the circular equation, then the value range of real number m is


According to the problem: (x-2m-1) ^ 2 + (y-m) ^ 2 + 4m ^ 2 + 5m + 2 - (2m + 1) ^ 2-m ^ 2 = 0
(x-2m-1)^2+(y-M)^2+4m^2+5m+2-4m^2-4m-1-m^2=0
(x-2m-1)^2-(y-m)^2=m^2-m-1
To make the equation a circle, i.e. m ^ 2-m-1 > 0
(m-1/2)^2>3/4
m> (3 ^ 0.5 + 1) / 2 or M



The equation x2 + Y2 + 2ay + 2A2 + A-1 = 0 represents a circle, then the value range of real number a is?


x²+(y+a)²=-a²-a+1
Represents a circle
So - A & sup2; - A + 1 > 0
a²+a-1