If the intersection point P of two straight lines y = x + 2k and y = 2x + K + 1 is inside the circle x ^ 2 + y ^ 2 = 4, the value of K is obtained Two linear equations are established, and x = k-1, y = 3K-1 are solved The center of the circle is at the origin, radius = 2 {-2

If the intersection point P of two straight lines y = x + 2k and y = 2x + K + 1 is inside the circle x ^ 2 + y ^ 2 = 4, the value of K is obtained Two linear equations are established, and x = k-1, y = 3K-1 are solved The center of the circle is at the origin, radius = 2 {-2

{-2

Given that the intersection P of two straight lines y = x + 2k and y = 2x + K + 1 is on the circle x2 + y2 = 4, then the value of K is ()
A. −15 ， −1B. −15 ， 1C. −13 ， 1D. -2，2

The intersection of two straight lines is the solution of the equation system y = x + 2ky = 2x + K + 1, where (x, y) = (k-1, 3K-1). The point is on the circle x2 + y2 = 4, if and only if (k-1) 2 + (3K-1) 2 = 4, the solution is k = 1, or K = − 15, so B

If the intersection of the line y = x + 2k and 4Y = 2x + 2K + 1 is in the circle x ^ 2 + y ^ 2 = 1, then the value range of K is?

Intersection: (0.5-3k, 0.5-k)
In a circle, x ^ 2 + y ^ 2

Find the minimum value of the center distance of two circles x ^ 2 + y ^ 2 + 2kx = k ^ 2-1 = 0, x ^ 2 + y ^ 2 + 2 (K + 1) y + K ^ 2 + 2K = 0!

The equation of circle: X & sup2; + Y & sup2; + 2kx + K & sup2; - 1 = 0 (x + k) & sup2; + Y & sup2; = 1 Center (- K, 0) x & sup2; + Y & sup2; + 2 (K + 1) y + K & sup2; + 2K = 0x & sup2; + (y + K + 1) & sup2; = 1 Center (0. - k-1) center distance = √ (- k-0) & sup2; + (0 + K + 1) & sup2; = √ (2k & sup2; + 2K +