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.
Best Answer
Draw the line passing through $(a,b)$ and the centre of the circle, meeting the circle at $P$ and $Q$. $P$ and $Q$ are the nearest and the farthest points on the circle from $(a,b)$. So they are the centres of the smallest and the largest circles. If $d$ is the distance between $(a,b)$ and the centre of the given circle. Then the radii of the smallest and the largest circles are respectively $d-3$ and $d+3$. So their difference is $6$.