$3.$ Equation of circle passing through $\left(-1,-2\right)$ and concentric with the circle $x^{2}+y^{2}+3x+4y+1=0$ $1\right)$ $x^{2}+y^{2}+3x+4y+6=0$ $2\right)$ $x^{2}+y^{2}-3x-4y+6=0$ $3\right)$ $x^{2}+y^{2}+3x+4y-6=0$ $4\right)$ $x^{2}+y^{2}+3x-4y-6=0$

Solve the system by using the inverse of the coefficient matrix. $3x+4y+z=6$ $3x-4y-z=-12$ $2x+y+3z=-8$

$x+y+z=3$ $x+2y+3z=4$ $x+4y+9z=6$ $m$ $4^{°}$ $41$

$6$ The circle cutting $x^{2}+y^{2}-6x+4y-12=0$ orthogonally and having centre $\left(-1.2\right)$ $15$ $1\right)$ $x^{2}+y^{2}+2x-4y-2=0$ $2\right)$ $x^{2}+y^{2}+2x-4y+2=0$ $3\right)$ $x^{2}+y^{2}-2x+4y-2=0$ $4\right)$ $x^{2}+y^{2}+2x-4y-4=0$

$Q2.$ If $P\left(a,b\right)$ lies on $2x-3y+2=0$ and $Q\left(b,a\right)$ lies on $\square $ $7x+2y-3=0$ then the equation of PQ is $\square $ O $1a\right)$ $9x-y-1=0$ $1\square $ $\left(n\right)$ $4x-4y-1=0$ $\left(c\right)$ $4x+4y=1$ $-$ $\left(D\right)$ $4\times +4y+1=0$

$-39x=\dfrac {1} {2}\left(x+y$ $13$ $1x+y+x=0$ show that $x^{2}+$ $+x^{2}=3$ $14$ With