Question

find the equation of the circle which touches both the axes at a distance of 6 units from the origin

Answer 1

Hi Mate!!!

Equation of circle is

( x - 6 )² + ( y - 6 ) ² = 36

Answer 2

$x^2+y^2 \pm 12x \pm 12y+36=0$ is the Required Equation

Step-by-step explanation:

As we know,

$x^2+y^2+2gx+2fy+c=0$. is the General equation of circle

Radius = $\sqrt {g^2+f^2-c}$ = 6 (here).

So, c = $g^2+f^2-36$

Rewriting the equation :

$x^2+y^2+2gx+2fy+(g^2+f^2-36)=0$

And Center = ($\pm 6,\pm 6$) (As per question)

So, (-g,-f) = ($\pm 6,\pm 6$)

replacing new found values of g,f in equation of circle is :

$x^2+y^2 \pm 12x \pm 12y+36=0$