Question

The equation of the circle whose radius is $4$, centre lies in the first quadrant and which touches $x$-axis and line $4x-3y=0$ is

• A. $x^2+y^2-4x-16y+4=0$
• B. $x^2+y^2-8x-4y+16=0$
• C. $x^2+y^2-16x-8y+64=0$
• D. $x^2+y^2-4x-8y+4=0$

Answer 1

Answer: (C)

$x^2+y^2-16x-8y+64=0$

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

Since it touches $x$-axis, the distance of the centre from the $x$-axis must be equal to the radius.

So, $-f=4$ ....($\because$ the circle lies in the first quadrant)

We know, $g^2+f^2-c=r^2$

Radius is $4$

$⟹g^2+16-c=16$

$⟹g^2=c$

Perpendicular distance of centre $(x_1,y_1)$ from line $ax+by+c=0$ is

$\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

Also, the perpendicular distance from centre $(-g,-f)$ to the line $4x-3y=0$ would be equal to its radius.

$⟹\frac{-4g+3f}{5}=\pm 4$

$⟹-4g-12=\pm 20$ ..... $(f=-4)$

$⟹g=2$ or $g=-8$

Since the circle lies in first quadrant, $g=-8$

$⟹c=64$ ........$(g^2=c)$

Hence, we have the equation of the circle as $x^2+y^2-16x-8y+64=0$.