Question

The equation of the circle of the radius $2\sqrt{2}$ whose centre lies on the line $x-y=0$ and which touches the line $x+y=4$, and whose centre's coordinates satisfy the inequality $x+y>4$ is

• A. $x^2+y^2-8x-8y+24=0$
• B. $x^2+y^2=8$
• C. $x^2+y^2-8x+8y=24$
• D. none of these

Answer 1

The correct option is A $x^2+y^2-8x-8y+24=0$

The lines $x-y=0$ and $x+y=4$ are perpendicular lines so the point of intersection is the point on contact of tangent and the circle is $(x,y)=(2,2)$

Let the coordinates of the centre be $(a,a)$ as the point lies on the line $x=y$.

We have the radius $=2\sqrt{2}$. So the distance between $(a,a)$ and $(2,2)$ is $2\sqrt{2}$

$\Rightarrow 2(a-2{)}^2=8$

$(a-2)=\pm 2$

$a=4$ and $a=0$

Possible points are $(0,0)$ and $(4,4)$ but as the centre points also satisfy the inequality $x+y>4$ .We have to consider the point $(4,4)$.

So the equation of the circle is $(x-4{)}^2+(y-4{)}^2=8$

$x^2+y^2-8x-8y+24=0$

Hence, option A is correct.