Question

If the line $ax+by+c=0$ touches the circle $x^2+y^2-2x=\frac{3}{5}$ and is normal to the circle $x^2+y^2+2x-4y+1=0$, then $(a,b)$ are

• A. $(1,3)$
• B. $(3,1)$
• C. $(1,2)$
• D. $(2,1)$

Answer 1

The correct option is B $(3,1)$

$x^2+y^2-2x=\frac{3}{5}\Rightarrow (x-1{)}^2+y^2=\frac{8}{5}$

So, Radius, $R=2\sqrt{\frac{2}{5}}$ and it's center is at $(1,0)$

ie, Distance, $d$ from the circle to $ax+by+c=0$ is,

$d=\frac{a\times 1+b\times 0+c}{\sqrt{a^2+b^2}}=\frac{a+c}{\sqrt{a^2+b^2}}=2\sqrt{\frac{2}{5}}⟶\left(1\right)$ (Inorder to satisfy the criterion of a tangent)

$x^2+y^2+2x-4y+1=0\Rightarrow (x+1{)}^2+(y-2{)}^2=4$

So, It's center is at $\left(\right(-1),2)$

As $ax+by+c=0$ is normal to the circle, it should go through the centre of the circle.

ie, $a-2b=c$ and $(y-2)=m(x+1)⟶\left(2\right)$

Substituting $c$ in (1),

$\frac{a+(a-2b)}{\sqrt{a^2+b^2}}=2\sqrt{\frac{2}{5}}$

$\Rightarrow \frac{a-b}{\sqrt{a^2+b^2}}=\sqrt{\frac{2}{5}}$

So, we can say $(a-b)=k\sqrt{2}$ and $a^2+b^2=5k^2$ foe some constant $k$.

$a^2+b^2-(a-b{)}^2=2ab=5k^2-2k^2=3k^2$

$(a-b{)}^2+4ab=(a+b{)}^2=6k^2+2k^2=8k^2\Rightarrow (a+b)=2k\sqrt{2}$

$a=\frac{1}{2}\left(\right(a+b)+(a-b\left)\right)=\frac{1}{2}\left(3k\sqrt{2}\right)$

$b=\frac{1}{2}\left(\right(a+b)-(a-b\left)\right)=\frac{1}{2}\left(k\sqrt{2}\right)$

Slope of the line, $m=\frac{dy}{dx}$

$\frac{d}{dx}(ax+by+c)=0\Rightarrow a+b\frac{dy}{dx}=0$

ie, $m=\frac{(-a)}{b}=(-3)$ (from above equations of $a$ and $b$)

Substituting the slope in (2),

$(y-2)=(-3)(x+1)\Rightarrow 3x+y+1=0$

Compairing with general equation given,

$(a,b)=(3,1)$

Option B is the correct answer.