The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is
AnswerCorrect option: D. $x^{2}+y^{2} \pm 6 x \pm 8 y=0$
(d) $x^{2}+y^{2} \pm 6 x \pm 8 y=0$
Explanation: Given that we need to find the equation of the circle passing through the origin and cuts off intercepts 6 and 8
from $x$ and $y$ - axes
Since the circle is having intercept a from $x$-axis the circle must pass through $(6,0)$ and $(-6,0)$ as it already passes through the origin.
Since the circle is having intercept 8 from $y$-axis the circle must pass through $(0,8)$ and $(0,-8)$ as it already passes through the origin.
Let us assume the circle passing through the points $\mathrm{O}(0,0), \mathrm{A}(6,0)$ and $\mathrm{B}(0,8)$.
We know that the standard form of the equation of the circle is given by:
$
\begin{equation*}
\Rightarrow x^{2}+y^{2}+2 f x+2 g y+c=0 \ \ldots (1)
\end{equation*}
$
Substituting $\mathrm{O}(0,0)$ in (1), we get,
$\Rightarrow 0^{2}+0^{2}+2 \mathrm{f}(0)+2 \mathrm{~g}(0)+\mathrm{c}=0$
$\Rightarrow \mathrm{c}=0$$\ldots(2)$
Substituting $A(6,0)$ in (1), we get,
$\Rightarrow 6^{2}+0^{2}+2 \mathrm{f}(6)+2 \mathrm{~g}(0)+\mathrm{c}=0$
$\Rightarrow 36+12 \mathrm{f}+\mathrm{c}=0$$\ldots(3)$
Substituting $B(0,8)$ in (1), we get,
$\Rightarrow 0^{2}+8^{2}+2 \mathrm{f}(0)+2 \mathrm{~g}(8)+\mathrm{c}=0$
$
\begin{equation*}
\Rightarrow 64+16 \mathrm{~g}+\mathrm{c}=0 \ldots (4)
\end{equation*}
$
On solving (2), (3) and (4) we get,
$\Rightarrow \mathrm{f}=-3, \mathrm{~b}=-4$ and $\mathrm{c}=0$
Substituting these values in (1), we get
$\Rightarrow x+y^{2}+2(-3) x+2(-4) y+0=0$
$\Rightarrow x^{2}+y^{2}-6 x-8 y=0$
Similarly, we get the equation $x^{2}+y^{2}+6 x+8 y=0$ for the circle passing through the points $(0,0),(-6,0),(0,-8)$, $x^{2}+y^{2}-6 x+8 y=0$ for the circle passing through the points $(0,0),(6,0),(0,-8)$.
$\therefore$ The equations of the circles are $\mathrm{x}^{2}+\mathrm{y}^{2} \pm 6 \mathrm{x} \pm 8 \mathrm{y}=0$