A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. Find the equation of the circle with center at $\left(6, 5\right)$ and touching the above circle externally.
Let the given circle (which is in the first quadrant), touch the X and Y axes at the points $A \left(a, 0\right)$ and $B \left(0, b\right)$ respectively.
Let C be the center of the given circle.
Then, $C = \left(a, b\right)$
Given: Radius of the circle $= 2$
i.e. $CA = CB = 2$
Now, $CA = \sqrt{\left(a - a\right)^2 + \left(b - 0\right)^2} = b$
and $\;$ $CB = \sqrt{\left(a - 0\right)^2 + \left(b - b\right)^2} = a$
$\therefore$ $\;$ $a = b = 2$
$\therefore$ $\;$ Center of the given circle $= C = \left(2, 2\right)$
Let the required circle touch the given circle at the point $P$.
Given: $\;$ center of the required circle $= C_1 = \left(6, 5\right)$
Radius of the given circle $= PC_1$
Now, $CC_1 = \sqrt{\left(6 - 2\right)^2 + \left(5 - 2\right)^2} = \sqrt{4^2 + 3^2} =5$
But $CC_1 = CP + PC_1$
$\therefore$ $\;$ $PC_1 = CC_1 - CP = 5 - 2 = 3$
$\therefore$ $\;$ Radius of the required circle $= 3$
$\therefore$ $\;$ Equation of the required circle is
$\left(x - 6\right)^2 + \left(y - 5\right)^2 = 3^2$
i.e. $x^2 + y^2 - 12 x - 10 y + 36 + 25 = 9$
i.e. $x^2 + y^2 - 12 x - 10 y + 52 = 0$