Question

The equation of the circle whose radius is $4$ and which touches the negative x-axis at a distance $3$ units from the origin is-

• A. $x^2+y^2-6x+8y-9=0$
• B. $x^2+y^2\pm 6x-8y+9=0$
• C. $x^2+y^2+6x\pm 8y+9=0$
• D. $x^2+y^2\pm 6x-8y-9=0$

Answer 1

Answer: (C)

$x^2+y^2+6x\pm 8y+9=0$

Given that, the radius of a circle is $4$ and it touches the negative x-axis at a distance of $3$ units.

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ----- (1)

As it touches x-axis, x-intercept $2\sqrt{g^2-c}=0$

$\Rightarrow g^2=c$

As $y=0$ is tangent to the circle, using the condition, the radius is equal to perpendicular distance from the center.

$|f|=4$

from (1) equation of circle is $x^2+y^2+2gx\pm 8y+g^2=0$

But, $(-3,0)$ lies on the circle.

$\Rightarrow 9-6g+g^2=(g-3{)}^2=0$

$\Rightarrow g=3$

$\therefore$ Required equation of circle is $x^2+y^2+6x\pm 8y+9=0$
Hence, option C.