JEE Main 2021 — Circle Question with Solution

Let a point circle: ${(x-1)}^2+{(y-2)}^2=0$ 

Equation of circle passing through the intersection of the line and the circle:$C_1+\lambda L_1=0$

$(x-1{)}^2+(y-2{)}^2+\lambda (4x+3y-10)=0$
$x^2+y^2+(2\lambda -1)2x+\left(\frac{3}{2}\lambda -2\right)2y+5-10\lambda =0$

$r=\sqrt{g^2+f^2-c}$
$r=\sqrt{(2\lambda -1{)}^2+{\left(\frac{3}{2}\lambda -2\right)}^2-(5-10\lambda )}=5$
$4{\lambda }^2+1-4\lambda +\frac{9}{4}{\lambda }^2+4-6\lambda -5+10\lambda =25$
$\frac{25}{4}{\lambda }^2-25=0$
$\lambda =\pm 2$
for $\lambda =2$
$x^2+y^2+6x+2y-15=0;C_1(-3,-1)$
for $\lambda =-2$
$x^2+y^2-10x-10y+25=0;C_2(5,5)$
$\left|\right(\alpha +\beta \left)\right(\gamma +\delta \left)\right|=40$