Find the equation of the circle which passes through the origin and cuts-off intercepts of lengths 6 and 8 units from the positive parts of X and Y axes respectively.
Let the required circle cut the X axis at point A and the Y axis at point B.
Given: $\;\;\;$ X intercept $= 6$ units and Y intercept $= 8$ units
$\therefore$ $\;$ $A = \left(6,0\right)$ and $B = \left(0,8\right)$
$OA$ and $OB$ are the chords of the circle.
Let $C$ be the center of the circle.
Now, perpendicular from the center to a chord bisects the chord.
Draw perpendiculars from center $C$ to chords $OA$ and $OB$.
Then $M$ and $N$ are the midpoints of $OA$ and $OB$ respectively.
$\therefore$ $\;$ $M = \left(3,0\right)$ and $N = \left(0,4\right)$
$\therefore$ $\;$ Center $C = \left(3,4\right)$
Radius of the circle $= OC = \sqrt{\left(3 - 0\right)^2 + \left(4 - 0\right)^2} = 5$
$\therefore$ $\;$ Equation of required circle is
$\left(x - 3\right)^2 + \left(y - 4\right)^2 = 5^2$
i.e. $x^2 + y^2 - 6x - 8y + 9 + 16 = 25$
i.e. $x^2 + y^2 - 6x - 8y = 0$