Multiple choice

The length of tangent from any point in the circle $x^2+y^2+4x-6y-12=0$ to the circle $x^2+y^2+4x-6y+4=0$ is?

  1. $2$
  2. $16$
  3. $8$
  4. $4$
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D Correct answer
Explanation

The length of the tangent from a point (x1, y1) to a circle S=0 is sqrt(S(x1, y1)). Here, the point is on the first circle, so S1(x1, y1) = 0. The length of the tangent to the second circle S2=0 is sqrt(S2(x1, y1)). S2 = x^2+y^2+4x-6y+4 = (x^2+y^2+4x-6y-12) + 16 = 0 + 16 = 16. The length is sqrt(16) = 4.

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