Question

A circle touches both the $x-$axis and the line $4x-3y+4=0$. Its centre is in the third quadrant and lies on the line $x-y-1=0$. Find the equation of the circle.

Answer 1

We have,

Line $4x-3y+4=0$….. (1)

Line $x-y-1=0$ …… (2)

From equation (1) and (2) to and we get,

$x=-7andy=-8$

Then,

Centre of circle $(h,k)=(-7,-8)$

But centre lies third quadrant.

So,

centre$(h,k)=(-7,8)$

so,

Equation of circle

${(x-h)}^2+{(y-k)}^2=r^2$

$\Rightarrow {(x+7)}^2+{(y-8)}^2={(-7)}^2(\therefore x-axisso,radiusofcircle=Abssicissa)$

$\Rightarrow x^2+49+14x+y^2+64-16y=49$

$\Rightarrow x^2+y^2+14x-16y+64=0$

Hence, the required equation of circle.