Question

Find the equation of the circle touching the negative y-axis at a distance $5$ form the origin and intercepting a length $8$ on the x-axis.

Answer 1

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ Since it touches y axis at (0, -3) and (0, -3) lies on the circle
$\therefore c=f^2$ .....(i)
9 - 6f + c = 0 .......(ii)
From (i) and (ii) we get 9 - 6f + $f^2$ = 0
$\Rightarrow {(f-3)}^2=0\Rightarrow f=3$
Putting f = 3 in (i) we obtain c = 9
It is given that the circle $x^2+y^2+2gx+2fy+c=0$ intercepts length 8 on x axis
$\therefore 2\sqrt{g^2-c}=8\Rightarrow 2\sqrt{g^2-9}=8\Rightarrow g^2-9=16\Rightarrow g=\pm 5$
Hence the required circle is $x^2+y^2\pm 10x+6y+9=0$