The circle touching the $y$-axis at a distance 4 units from the origin and cutting off an intercept 6 from…
The circle touching the $y$-axis at a distance 4 units from the origin and cutting off an intercept 6 from $\mathrm{x}$-axis is
- $x^2+y^2 \pm 10 x-8 y+16=0$
- $x^2+y^2 \pm 5 x-8 y+16=0$
- $x^2+y^2 \pm 5 x-2 y-8=0$
- $x^2+y^2 \pm 2 x-y-12=0$
Solution
The given situation is possible with the 2 circles.
Considering centre at $C_1$
$x$ intercept $=\mathrm{PQ}=6$
$y$ intercept $=0$
for $x^2+y^2+2 g x+2 f y+c=0$ ...(i)
$x$ intercept $=2 \sqrt{g^2-c}=6$
$g^2-c=9$
$y$ intercept $=2 \sqrt{f^2-c}=0$
$f^2=\mathrm{c}$
$\begin{aligned} & \mathrm{PB}=1 / 2 \mathrm{PQ}=3 \\ & \mathrm{OA}=\mathrm{C}_1 \mathrm{~B}=4\end{aligned}$
In $\Delta \mathrm{p} C_1 \mathrm{~B}$,
$\mathrm{PC}_1==\sqrt{C_1 B^2+p B^2}$
$=\sqrt{3^2+4^2}$
$\mathrm{PC}_1=5$
radius of circle $=5$
radius $=\sqrt{g^2+f^2-c}$
$g^2+f^2-c=25$
$\begin{aligned} & f^2=c \\ & g^2=25 \\ & g= \pm 5 \\ & g^2-c=9 \\ & f^2+9=25 \\ & f^2=16 \\ & f= \pm 4\end{aligned}$
and $c=f^2=16$
Equation of circle is
$x^2+y^2 \pm 10 x \pm 8 y+16=0$
In given option correct is
$x^2+y^2 \pm 10 x-8 y+16=0$
Asked in: AP EAMCET 2022 (05 Jul Shift 2)
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