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The figure has the smallest circle on the left and the largest circle around the circle in the middle, which is our given circle$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.

Saradamani
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    Why does the radius of the largest circle equal to $6$ if you don't have the values of $a$ and $b$? – CY Aries Apr 06 '18 at 17:32
  • If you draw the circle $(x+1)^2+(y+2)^2=9$ and want the largest circle whose center is on the circumference of $(x+1)^2+(y+2)^2=9$, then it must have a radius $=6$. Am I wrong? – Saradamani Apr 06 '18 at 17:34
  • Think of a point $(a,b)$ which is very far away from $(-1,-2)$. The radius of the circle can be very large. – CY Aries Apr 06 '18 at 17:36
  • No but the circle we are talking about must have its center on the circumference of $(x+1)^2+(y+2)^2=9$ and hence there must be a bound. Oh no no no it can be so that the largest circle has a center on the circumference but does not touch the circle $(x+1)^2+(y+2)^2=9$ which I have wrongly assumed. Thanks for citing. – Saradamani Apr 06 '18 at 17:38

2 Answers2

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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 $6$.

CY Aries
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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, $$r\le\sqrt{(b+2)^2+(a+1)^2}+3\text { for } \sin\left(t+\arctan\dfrac{a+1}{b+2}\right)=-1$$

lab bhattacharjee
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