Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (–4, 1) and having their centers on the circumference of the circle x2+y2+2x+4y-4=0. If r1r2=a+b2, then a+b is equal to :

# Let ${r}_{1}$ and ${r}_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point (–4, 1) and having their centers on the circumference of the circle ${x}^{2}+{y}^{2}+2x+4y-4=0.$ If $\frac{{r}_{1}}{{r}_{2}}=a+b\sqrt{2}$, then $a+b$ is equal to :

1. A

3

2. B

7

3. C

5

4. D

11

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### Solution:

The given circle is ${x}^{2}+{y}^{2}+2x+4y-4=0$

The center and radius of the circle are $C\left(-1,-2\right)$ and $r=\sqrt{1+4+4}=3$

Given  ${r}_{1}$ and ${r}_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point $P\left(-4,1\right)$ and having their centers on the circumference of the circle ${x}^{2}+{y}^{2}+2x+4y-4=0.$

Here ${r}_{1}=CP+r=3\sqrt{2}+3$ and ${r}_{2}=CP-r=3\sqrt{2}-3$

Consider $\frac{{r}_{1}}{{r}_{2}}=a+b\sqrt{2}$

Hence,

$\begin{array}{rcl}\frac{3\sqrt{2}+3}{3\sqrt{2}-3}& =& a+b\sqrt{2}\\ \frac{18+9+18\sqrt{2}}{9}& =& a+b\sqrt{2}\\ 3+2\sqrt{2}& =& a+b\sqrt{2}\end{array}$

Therefore, $a+b=3+2=\overline{)\mathbf{5}}$

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