Question

Find the equation of circle which touched the x - axis and the line 4x - 3y + 4 = 0 It centre lying in the third quadrant and lies on the line x - y - 1 = 0

Answer 1

Let the circle beg, f, c so that its centre (- g, - f) is in 3rd quadrant so that both g and fare to be + ive. Centre lies on x - y - 1 = 0
$\therefore$ -g + f 1 = 0 .(1)
Touches x-axis
$\therefore I_x=2\sqrt{g^2-c}=0\therefore g^2=c$ ..(2)
Touches 4x 3y + 4 = 0 $\therefore$ p = r
$\Rightarrow \frac{-4g+3f+4}{\pm 5}=\sqrt{g^2+f^2-c}=f$, by (2)
$\therefore$ -4g + 3f + 4 = 5f or -5f
$\therefore$ -g + 2f + 1 = 0 ..(3)
or 2g + f 2 = 0.....(4)
Solving (1), (3) or (1), (4), we get
g = -3, f = -2 rejected
$g=\frac{1}{3},f=\frac{4}{3}$ both +ive $\therefore c=g^2=\frac{1}{9}$
$\therefore x^2+y^2+\frac{2}{3}x+\frac{8}{3}f+19=0$
or 9$(x^2+y^2)$ + 6x + 24y + 1 = 0.