Question

The locus of the point from which the length of the tangent to the circle $x^2+y^2-2x-4y+4=0$ is $3$ units is

• A. $x^2+y^2-2x-4y-9=0$
• B. $x^2+y^2-2x-4y-4=0$
• C. $x^2+y^2-2x-4y-3=0$
• D. $x^2+y^2-2x-4y-5=0$

Answer 1

Answer: (D)

$x^2+y^2-2x-4y-5=0$

Let the point is $(h,k)$.

Then the length of the tangent to the circle $x^2+y^2-2x-4y+4=0$ is $\sqrt{h^2+k^2-2h-4k+4}$ units.

Now according to the given problem we've,

$\sqrt{h^2+k^2-2h-4k+4}=3$

or, $h^2+k^2-2h-4k+4=9$

or, $h^2+k^2-2h-4k-5=0$.

So the locus of $(h,k)$ is $x^2+y^2-2x-4y-9=0$.