A common tangent to the circle x^2+y^2=9 and | vedclass

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(B) Let $y=mx+c$ be the equation of a common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=9$. Condition for $y=mx+c$ to be a tangent to th

Vedclass · Accepted Answer

$x-\sqrt{3}y+6=0$

Vedclass · Answer

(B) Let $y=mx+c$ be the equation of a common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=9$. Condition for $y=mx+c$ to be a tangent to the parabola $y^2=4ax$ is $c=\frac{a}{m}$. Here $4a=8$,so $a=2$. Thus,$c=\frac{2}{m}$ $(i)$. The line $y=mx+c$ is also tangent to the circle $x^2+y^2=9$,so the perpendicular distance from the center $(0,0)$ to the line $mx-y+c=0$ must equal the radius $r=3$. $\frac{|c|}{\sqrt{m^2+1}}=3 \Rightarrow c^2=9(m^2+1)$. Substituting $c=\frac{2}{m}$ into the equation: $\frac{4}{m^2}=9(m^2+1)$ $\Rightarrow 4=9m^2(m^2+1)$ $\Rightarrow 9m^4+9m^2-4=0$. Let $m^2=t$,then $9t^2+9t-4=0 \Rightarrow (3t-1)(3t+4)=0$. Since $m^2=t > 0$,we have $t=\frac{1}{3}$,so $m^2=\frac{1}{3}$ and $m=\pm\frac{1}{\sqrt{3}}$. For $m=\frac{1}{\sqrt{3}}$,$c=\frac{2}{1/\sqrt{3}}=2\sqrt{3}$. The equation is $y=\frac{1}{\sqrt{3}}x+2\sqrt{3}$ $\Rightarrow \sqrt{3}y=x+6$ $\Rightarrow x-\sqrt{3}y+6=0$.

$A$ common tangent to the circle $x^2+y^2=9$ and parabola $y^2=8x$ is

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