Question

If the circle $x^2+y^2-4x-8y+16=0$ rolls up (away from $x-$axis) along the tangent to it at $(2+\sqrt{3},3)$ by $2$ units and its equation in the new position is $x^2+y^2+2ax+2by+c=0,$ then the value of $a^2+8b+c$ is equal to

Answer 1

Given circle : $x^2+y^2-4x-8y+16=0$
Equation of tangent at $(2+\sqrt{3},3)$ is $T=0$
$\Rightarrow x(2+\sqrt{3})+y\left(3\right)-2(x+2+\sqrt{3})-4(y+3)+16=0$
$\Rightarrow \sqrt{3}x-y=2\sqrt{3}$
$\Rightarrow$ Slope of tangent $=\sqrt{3}$
$\Rightarrow$ Inclination $={60}^{\circ }$

Let $C_2$ be the new centre after rolling.
Original centre $C_1\equiv (2,4)$
and $C_1{C}_2=2$ (as new centre is $2$ units away from $C_1$ along the tangent)
$\therefore C_2\equiv (2+2\cos {60}^{\circ },4+2\sin {60}^{\circ })\equiv (3,4+\sqrt{3})$
$\therefore$ Equation of circle in new position is : $(x-3{)}^2+(y-4-\sqrt{3}{)}^2=2^2$
$\Rightarrow x^2+y^2-6x-2(4+\sqrt{3})y+24+8\sqrt{3}=0$ Comparing with $x^2+y^2+2ax+2by+c=0,$
$a=-3,b=-(4+\sqrt{3}),c=24+8\sqrt{3}$
$a^2+8b+c=1$