Question

the equation of the circle which touches the line $x=0$ & $y=0$ & $3x+4y=4$

Answer 1

Let's take the centre of circle as $(h,k)$. As given circle touches both x and y axis so we can conclude $h=k$ means centre of the circle is at $(h,h)$

Now it touches the line $3x+4y-4=0$

Hence perpendicular distance from centre on the line is radius which $h$ as per the condition so, we get an equation $\frac{3h+4h-4}{\sqrt{3^2+4^2}}=h$

So, $7h-4=5h$ and we get $h=2$

So centre at $(x,2)$

Radius $=h=2$

Now from the definition of circle equation is

$(x-2{)}^2+(y-2{)}^2=4$

And by solving we will get equation as

$x^2+y^2-4x-4y+4=0$