<s>[INST] The difference between the radii of the largest and smallest circles having centres on the circumference of $x^2+2x+y^2+4y=4$ Owing to restriction of 150 characters in the title section I include the latter part of the problem here below in bold and italics
Also given that both the circles(largest and smallest) pass through a point $(a,b)$ lying outside the given circle, $x^2+2x+y^2+4y=4$
My attempt:
$x^2+2x+y^2+4y=4$
$\implies\space (x+1)^2+(y+2)^2=(3)^2$
I have drawn a circle with center $(-1,-2)$
& Also the largest circle with a center at the circumference of $x^2+2x+y^2+4y=4$ is the circle with $radius=6$
Also this circle passes through $(a,b)$
Now the smallest circle with center on the circumference of $x^2+2x+y^2+4y=4$ and passing through $(a,b)$, let us assume, has its center $(\alpha,\beta)$  and radius $=r$.
Thus $(\alpha-a)^2+(\beta-b)^2=r^2\cdot\cdot\cdot(1)$
Again if the largest circle has radius $R=6$
Then it has its center(by observation) at $(2,-2)$
Thus Equation of the largest circle is $(x-2)^2+(y+2)^2=36$
Again this largest circle passes through $(a,b)$ 
$\therefore$ $(a-2)^2+(b+2)^2=36  \cdot\cdot\cdot(2)$
I am stuck here. Please throw some light.
 [/INST] Any point on the circle $$(x+1)^2+(y+2)^2=3^2$$ can be $P(3\cos t-1,3\sin t -2)$
So, if $r$ is the radius of the new circle $$r^2=(a+1-3\cos t)^2+(b+2-3\sin t)^2$$
$$=(a+1)^2+(b+2)^2+9-6\{(b+2)\sin t+(a+1)\cos t\}$$
$$=(a+1)^2+(b+2)^2+9-6\sqrt{(b+2)^2+(a+1)^2}\sin\left(t+\arctan\dfrac{a+1}{b+2}\right)$$
$$r\ge|\sqrt{(b+2)^2+(a+1)^2}-3|\text { for } \sin\left(t+\arctan\dfrac{a+1}{b+2}\right)=1$$
If $(a,b)$ resides outside $$(x+1)^2+(y+2)^2=3^2,(a+1)^2+(b+2)^2>3^2,$$
$$\implies|\sqrt{(b+2)^2+(a+1)^2}-3|=+(\sqrt{(b+2)^2+(a+1)^2}-3)$$
Similarly, $$r\le\sqrt{(b+2)^2+(a+1)^2}+3\text { for } \sin\left(t+\arctan\dfrac{a+1}{b+2}\right)=-1$$
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