The locus of a point which is at a distance of 2 units from the line 2x-3y+4=0 and at a distance of $\sqrt{13} \mathrm{units} \mathrm{from} \mathrm{a} \mathrm{point} (5,0), \mathrm{is}$

$P\left(x_1,y_1\right)$ be a ;ocus on the point

2 units distance from the point $P\left(x_1,y_1\right)$ to the line 2x-3y+4=0 is

$\frac{\left|2x_1-3y_1+4\right|}{\sqrt{4+9}}=2$

s.o.b.s

${\left(2x_1-3y_1+4\right)}^2=52···\left(i\right)$

and distance between $P\left(x_1,y_1\right)$ to the point $\left(5,0\right)$ is $\sqrt{13}$

$\sqrt{{\left(x_1-5\right)}^2+{\left(y_1-0\right)}^2}=\sqrt{13}$

s.o.b.s

$\left({x_1}^2+25-10x_1+{y_1}^2\right)=13···\left(ii\right)$

from $\left(i\right)$ and $\left(ii\right)$ will get,

$12x_1{y}_1-5{{\mathrm{y}}_1}^2-56{\mathrm{x}}_1+24{\mathrm{y}}_1+84=0$

locus is

$12\mathrm{xy}-5{\mathrm{y}}^2-56\mathrm{x}+24\mathrm{y}+84=0$