Let r₁ and r₂ be the radii of the largest and smallest circles, respectively, which pass through the point (−4, 1) and having their centres on the circumference of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}+4\mathrm{y}-4=0$. If $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\mathrm{a}+\mathrm{b}\sqrt{2}$, then a + b is equal to

Given, circle $\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}+4\mathrm{y}-4=0$

$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+2\mathrm{x}+1\right)+\left({\mathrm{y}}^2+4\mathrm{y}+4\right)-1-4-4=0\\ \Rightarrow (\mathrm{x}+1{)}^2+(\mathrm{y}+2{)}^2=9\end{array}$

Any point on the given circle be

$\left[\right(3\cos \mathrm{\theta }-1),(3\sin \mathrm{\theta }-2\left)\right]$

[to find the point coordinates, take x + 1 = 3 cos $\mathrm{\theta }$ and y + 2 = 3 sin $\mathrm{\theta }$]

Now, circle passes through (-4, 1) and their centres lie on the given circle.

So, the centre coordinate of that circle be $(3\cos \mathrm{\theta }-1,3\sin \mathrm{\theta }-2)$.

Since, it passes through (-4, 1), then radius of this circle be

$\begin{array}{c}\mathrm{r}=\sqrt{(3\cos \mathrm{\theta }-1+4{)}^2+(3\sin \mathrm{\theta }-2-1{)}^2}\\ =\sqrt{9{\cos }^2\mathrm{\theta }+9+18\cos \mathrm{\theta }+9{\sin }^2\mathrm{\theta }+9-18\sin \mathrm{\theta }}\\ =\sqrt{27+18(\cos \mathrm{\theta }-\sin \mathrm{\theta })}\\ =3\sqrt{3+2(\cos \mathrm{\theta }-\sin \mathrm{\theta })}\end{array}$

Maximum radius will be when $(\cos \mathrm{\theta }-\sin \mathrm{\theta })$ is maximum i.e.

$\begin{array}{l}\cos \mathrm{\theta }-\sin \mathrm{\theta }=\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}\\ \therefore {\mathrm{r}}_1={\mathrm{r}}_{max}=3\sqrt{3+2/\sqrt{2}}\end{array}$

Minimum radius will be when $(\cos \mathrm{\theta }-\sin \mathrm{\theta })$ is minimum i.e. $-\sqrt{2}$.

$\therefore {\mathrm{r}}_2={\mathrm{r}}_{min}=3\sqrt{3-2\sqrt{2}}$

Given, $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\mathrm{a}+\mathrm{b}\sqrt{2}\text{, then }\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}=(\mathrm{a}+\mathrm{b}\sqrt{2}{)}^2$

$\begin{array}{l}\Rightarrow \frac{9(3+2\sqrt{2})}{9(3-2\sqrt{2})}=(\mathrm{a}+\mathrm{b}\sqrt{2}{)}^2\\ \Rightarrow \frac{(3+2\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}=(\mathrm{a}+\mathrm{b}\sqrt{2}{)}^2\\ \Rightarrow \frac{(3+2\sqrt{2}{)}^2}{1}=(\mathrm{a}+\mathrm{b}\sqrt{2}{)}^2\end{array}$

Comparing coefficients, a = 3, b = 2

$\therefore$ a + b = 5