If the tangent to the circle $x^2+y^2-4 x+2 y-5=0$ at $(3,-4)$ cuts the circle $x^2+y^2+16 x+2 y+10=0$ at…

If the tangent to the circle $x^2+y^2-4 x+2 y-5=0$ at $(3,-4)$ cuts the circle $x^2+y^2+16 x+2 y+10=0$ at $A$ and $B$, then the mid point of $A B$ is:
  1. $(-6,-9)$
  2. $(-9,-6)$
  3. $(-6,-7)$
  4. $(-7,-6)$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2018 (24 Apr Shift 2)

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