i) If the straight line 2x + 3y = 1 intersect the circle x² + y² = 4 at the points A and B, then find the equation of the circle having AB as diameter.

Given circle is $\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-4=0$
Given line is $\mathrm{L}=2\mathrm{x}+3\mathrm{y}-1=0$
circle S = 0 and the line  L = 0 Intersect at A,B.
Required circle passes through the points of intersection of A and B.
$\therefore \text{ Equation of required circle is }\mathrm{S}+\mathrm{\lambda L}=0$
$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4+\mathrm{\lambda }(2\mathrm{x}+3\mathrm{y}-1)=0 ...\left(1\right)$
$\text{ Its centre C}=\left(-\frac{2\mathrm{\lambda }}{2},-\frac{3\mathrm{\lambda }}{2}\right)=\left(-\mathrm{\lambda },-\frac{3}{2}\mathrm{\lambda }\right)$

$\because$AB is diameter, centre  C lies on it i.e., $\mathrm{c}\left(-\mathrm{\lambda },\frac{-3}{2}\mathrm{\lambda }\right)$ lies on line
$\begin{array}{l}2\mathrm{x}+3\mathrm{y}-1=0\\ \therefore 2(-\mathrm{\lambda })+3\left(-\frac{3}{2}\mathrm{\lambda }\right)-1=0\\ \frac{-4\mathrm{\lambda }-9\mathrm{\lambda }-2}{2}=0\Rightarrow -13\mathrm{\lambda }-2=0\Rightarrow \mathrm{\lambda }=-\frac{2}{13}\end{array}$
Put $\lambda$ in (1)

 Equation of required circle is
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-4-\frac{2}{13}(2\mathrm{x}+3\mathrm{y}-1)=0\\ \Rightarrow 13\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)-52-4\mathrm{x}-6\mathrm{y}+2=0\\ \Rightarrow 13\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)-4\mathrm{x}-6\mathrm{y}-50=0\end{array}$