Question

The equation of the circle that touches the $Y$ - axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X$ - axis is

• A. $x^{2}+y^{2} \pm 5 x-8 y+16=0$
• B. $x^{2}+y^{2} \pm 2 x-4 y=0$
• C. $x^{2}+y^{2} \pm 3 x-2 y-8=0$
• D. $x^{2}+y^{2} \pm 10 x-8 y+16=0$

Answer 1

Answer: (D)

Equation of circle touches the $Y$ -axis at a distance 4 unit from origin

$(x-h)^{2}+(y-k)^{2}=r^{2}$
$O A^{2}=O C^{2}+A C^{2}$
$O A^{2}=4^{2}+3^{2} \quad[A B=6]$
$O A^{2}=25=5^{2}$
$\therefore $ Equation of circle
$(x-h)^{2}+(y-4)^{2}=25$
Its also passes through $(0,4)$
$\because h^{2}=25 \Rightarrow h=\pm 5$
$\therefore $ Equation of circle
$(x \pm 5)^{2}+(y-4)^{2}=25$
$x^{2}+y^{2} \pm 10 x-8 y+16=0$