The length of the tangent drawn from a point on the circle $x^{2} + y^{2} - 4x + 6y - 4 = 0$ to the circle $x^{2} + y^{2} - 4x + 6y = 0$ is
- $8$
- $4$
- $2$
-
None of these
Circle 1: x^2 + y^2 - 4x + 6y - 4 = 0. Circle 2: x^2 + y^2 - 4x + 6y = 0. These are concentric circles with centers at (2, -3). The radius of circle 1 is sqrt(4 + 9 + 4) = sqrt(17). The radius of circle 2 is sqrt(4 + 9) = sqrt(13). The length of the tangent from a point on circle 1 to circle 2 is sqrt(R^2 - d^2), where d is the distance between centers (0 here) and R is the radius of the outer circle. This is not correct. The tangent from a point on C1 to C2 is sqrt(dist(point, center)^2 - r2^2). Since the point is on C1, dist^2 = r1^2 = 17. Length = sqrt(17 - 13) = sqrt(4) = 2.
The length of a tangent from a point on one circle to another is the square root of the difference of the two circle equations evaluated at that point. Let S1 be x squared plus y squared minus 4x plus 6y minus 4 equals 0 for the first circle, and S be x squared plus y squared minus 4x plus 6y equals 0 for the second circle. Subtracting the equations shows that S equals S1 plus 4, and since S1 equals 0 for points on the first circle, S equals 4. The tangent length is the square root of 4, which is 2.