Question

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

• A. $8$
• B. $4$
• C. $2$
• D. None of these

Answer 1

Answer: (C)

$2$

The equation of given circles is

$x^2+y^2-4x+6y-4=0$ and $x^2+y^2-4x+6y=0$

i.e. $(x-2{)}^2+(y+3{)}^2=17$ ..... $\left(i\right)$ and

$(x-2{)}^2+(y+3{)}^2=13$ .... $\left(ii\right)$

The center of both the circles is $(2,-3)$ and radii of first and second circle are $\sqrt{17},\sqrt{13}$ respectively

We can see, $x$ will be the length of required tangent,

Now, $17=x^2+13$

$⟹x=2$

Hence, option C is correct.