Question

Find the equation of circles which touches $x-$axis at $(4,0)$ and makes an intercept at $y-$axis of $6$ units.

Answer 1

Let the equation of circle is

$x^2+y^2+2gx+2fy+c=0$ ___ (1)

Given eq (1) touches x-axis

$\therefore c=g^2$ ___ (2)

point $(4,0)$ lies on eq (1)

$x=4,y=0$

$16+8g+g^2=0$

$g^2+2.4g+4^2=0$

$(g+4{)}^2=0$

$g+4=0$

$g=-4$

substitute $g=-4$ in (2)

$c=(-4{)}^2=16$

circle (1) cuts an intercept

of 6 units on y-axis

$\therefore 2\sqrt{f^2-c}=6$

$2\sqrt{f^2-16}=6$

$\sqrt{f^2-16}=3\Rightarrow f^2-16=9$

$f^2=25\Rightarrow f=\pm 5$

$\therefore$ The required equation of

circle is

$x^2+y^2-8x\pm 10y+16=0$