Distance between the curves from a particular point

Question

The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to

(A) 1

(B) 2

(C) 3

(4) 4

My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.

The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$

$D^2=(u-2)^2+v^2$

Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$

How we will proceed from here

Answer 1

The distance between the circle and $P(u,v)$ is the distance between $P$ and the centre minus the radius. So we have$$\left|\sqrt{(u+2)^2+v^2}-2\right|=\sqrt{(u-2)^2+v^2}$$Squaring,$$\begin{align*}&(u+2)^2+v^2+4-4\sqrt{(u+2)^2+v^2}=(u-2)^2+v^2\\\implies&2u+1=\sqrt{(u+2)^2+v^2}\\\implies&4u^2+4u+1=(u+2)^2+v^2\\\implies&u^2-v^2/3=1.\end{align*}$$

Answer 2

Distance $D_1$ of P(u,v) from the circle is $D_1=\sqrt{(u+2)^2+v^2}-2=\sqrt{F}-2$ and $D_2=\sqrt{(u-2)+v^2}=\sqrt{G}$, we require $$D_1=D_2 \implies \sqrt{(u+2)^2+v^2}-2=\sqrt{(u-2)+v^2} \implies \sqrt{F}-\sqrt{G}=2~~~(1)$$ We also have $$F-G=8u \implies \sqrt{F}+\sqrt{G}=4u~~~~(2)$$ From (1) and (2), we have $$\sqrt{F}=1+2u \implies u^2+4u+4+qu^2-q=1+4u+4u^2 \implies (q-3) u^2=(q-3)\implies q=3$$