Question

A circle touches both the x-axis and the line $4x-3y+4=0$. If its center is in the third quadrant and lies on the line $x-y-1=0$, then the equation of the circle is

• A. $9(x^2+y^2)+6x+24y-1=0$
• B. $9(x^2+y^2)+6x-24y+1=0$
• C. $9(x^2+y^2)+6x+24y+1=0$
• D. None of these

Answer 1

Answer: (C)

$9(x^2+y^2)+6x+24y+1=0$

Let the equation of the required circle be

$x^2+y^2+2gx+2gf+c=0$ ...(1)

Its centre is $C(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}.$ Since circle (1) touches the x-axis

$\therefore g^2-c=0$ or $c=g^2$ ...(2)

Again, since circle (1) touches the line

$4x-3y+4=0$

$\therefore \frac{|-4g+3f+4|}{5}=\sqrt{g^2+f^2-c}=\sqrt{f^2}=\left|f\right|$ [from (2)]

or, $-4g+3y+4=\pm 5f$

$\therefore 4g+2f=4$ or $2g+f=2$ ...(4)

$-4g+8f=-4$ or $g-2f=1$ ...(5)

Again, since center $C(-g,-f)$ lies on the line

$x-y-1=0$

$\therefore -g+f=1$ ...(6)

solving (4) and (6), we get $g=\frac{1}{3},f=\frac{4}{3}.$

Thus, $C\equiv (\frac{-1}{3},\frac{-4}{3})$ which lies in the third quadrant.

Also, from (2), $c=g^2=\frac{1}{9}.$

Solving (5) and (6), we get $f=-2,g=-3$

$\therefore C\equiv (3,2)$ which lies in the first quadrant.

Thus, for the required circle $g=\frac{1}{3},f=\frac{4}{3},c=\frac{1}{9}.$

$\therefore$ Equation of the required circle is

$x^2+y^2+\frac{2}{3}x+\frac{8}{3}y+\frac{1}{9}=0$

or, $9(x^2+y^2)+6x+24y+1=0.$