Question

The circle $x^2+y^2-4x-8y+16=0$ rolls up the tangent to it at $(2+\sqrt{3},3)$ by $2$ units, assuming the x-axis as horizontal, the equation of the circle in the new position is

• A. $x^2+y^2-6x-2(4+\sqrt{3})y+24+8\sqrt{3}=0$
• B. $x^2+y^2+6x-2(4+\sqrt{3})y+24+8\sqrt{3}=0$
• C. $x^2+y^2-6x+2(4+\sqrt{3})y+24+8\sqrt{3}=0$
• D. None of these

Answer 1

The correct option is A $x^2+y^2-6x-2(4+\sqrt{3})y+24+8\sqrt{3}=0$

Given circle is

$x^2+y^2-4x-8y+16=0$ ...(1)

Let $p=(2+\sqrt{3},3).$

Equation of tangent to the circle (1) at $p(2+\sqrt{3},3)$ is $(2+\sqrt{3})x-3y-2(x+2+\sqrt{3})-4(y+3)+16=0$

or, $\sqrt{3}x-y-2\sqrt{3}=0$ ...(2)

If line (2) makes an angle $\theta$ with the positive direction of x-axis, then $\tan \theta =\sqrt{3},$

$\therefore \theta ={60}^0.$

Let $A$ and $B$ be the centres of the in old and new positions, respectively, then

$A\equiv (2,4)$ and $B\equiv (2+2\cos {60}^0,4+2\sin {60}^0)(\because AB=2)$

Thus, $B\equiv (3,4+\sqrt{3}).$

Radius of the circle $=\sqrt{2^2+4^2-16}=2.$

$\therefore$ Equation of the circle in the new position is

${(x-3)}^2+{(y-4-\sqrt{3})}^2=2^2$

or, $x^2+y^2-6x-2(4+\sqrt{3})y+24+8\sqrt{3}=0$