A circle has radius 3 and its center lies on the line $y = x - 1$. Find the equation of the circle if it passes through the point $\left(7,3\right)$.
Let the center of the circle be $= C = \left(h,k\right)$
The center lies on the line $\;\;\;$ $y = x - 1$
$\therefore$ $\;$ We have, $\;\;\;$ $k = h - 1$ $\;\;\; \cdots \; (1)$
Given: Radius of required circle $= 3$
$\therefore$ $\;$ Equation of required circle is
$\left(x - h\right)^2 + \left(y - k\right)^2 = 3^2$
$\therefore$ $\;$ We have by equation $(1)$
$\left(x - h\right)^2 + \left(y - h + 1\right)^2 = 9$ $\;\;\; \cdots \; (2)$
The required circle passes through $\left(7,3\right)$.
$\therefore$ $\;$ We have from equation $(2)$,
$\left(7 - h\right)^2 + \left(3 - h + 1\right)^2 = 9$
i.e. $49 + h^2 - 14 h + 16 + h^2 - 8 h = 9$
i.e. $2h^2 - 22h + 56 = 0$
i.e. $h^2 - 11h + 28 = 0$
i.e. $\left(h - 7\right) \left(h - 4\right) = 0$
i.e. $h = 7$ $\;$ or $\;$ $h = 4$
Substituting the value of $h$ in equation $(1)$ gives
when $h = 7$, $k = 6$ and
when $h = 4$, $k = 3$
$\therefore$ $\;$ The centers of the required circles are $\left(7, 6\right)$ and $\left(4, 3\right)$.
$\therefore$ $\;$ The equations of the required circles are:
when center is $\left(7, 6\right)$:
$\left(x - 6\right)^2 + \left(y - 7\right)^2 = 9$
i.e. $x^2 + y^2 - 12 x - 14 y + 36 + 49 = 9$
i.e. $x^2 + y^2 - 12 x - 14 y + 76 = 0$ $\;\;\; \cdots \; (3)$
when center is $\left(4, 3\right)$:
$\left(x - 4\right)^2 + \left(y - 3\right)^2 = 9$
i.e. $x^2 + y^2 - 8 x - 6 y + 16 + 9 = 9$
i.e. $x^2 + y^2 - 8 x - 6 y + 16 = 0$ $\;\;\; \cdots \; (4)$
Equations $(3)$ and $(4)$ are the required equations of the circles.