Find the equations to the lines through the point $\left(1, 1\right)$ which are $3$ units distant from the point $\left(-2, 3\right)$.
Let the equation of the required line be $\;$ $Ax + By + C = 0$ $\;\;\; \cdots \; (1)$
Distance of the required line from the point $\left(x_1, y_1\right) = \left(-2, 3\right)$ $\;$ is $\;$ $= 3$ units
i.e. $\;$ $\dfrac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} = 3$
i.e. $\;$ $-2 A + 3 B + C = 3 \sqrt{A^2 + B^2}$ $\;\;\; \cdots \; (2)$
Since the required equation passes through the point $\left(1,1\right)$, we have
$A + B + C = 0$ $\implies$ $C = - A - B$ $\;\;\; \cdots \; (3)$
Substituting the value of $C$ in equation $(2)$ gives
$- 2A + 3B - A - B = 3 \sqrt{A^2 + B^2}$
i.e. $\;$ $-3A + 2B = 3 \sqrt{A^2 + B^2}$
i.e. $\;$ $9A^2 + 4B^2 -12AB = 9A^2 + 9B^2$
i.e. $\;$ $5B^2 = - 12 AB$
i.e. $\;$ $B \left(5 B + 12 A\right) = 0$ $\;\;\; \cdots \; (4)$
i.e. $\;$ $B = 0$ $\;\;\;$ OR $\;\;\;$ $5B + 12A = 0$
When $\;$ $5B + 12A = 0$ $\implies$ $A = \dfrac{-5}{12} B$
Substituting the value of $A$ in equation $(3)$ gives
$C = \dfrac{5}{12} B - B = \dfrac{-7}{12} B$
Substituting the values of $A$ and $C$ in equation $(1)$ gives
$\dfrac{-5}{12} B x + B y - \dfrac{7}{12} B = 0$
i.e. $\;$ $-5 x + 12 y - 7 = 0$
$\therefore \;$ The equation of the required line is $\;\;$ $5x - 12 y + 7 = 0$
When $\;$ $B = 0$, $\;$ we have from equation $(3)$, $\;\;\;$ $C = -A$
Substituting $C = -A$ and $B = 0$ in equation $(1)$ gives
$Ax - A = 0$
$\therefore \;$ The equation of the required line is $\;\;$ $x - 1 = 0$