Question

The length of common chord of circles $x^2+y^2+2x+6y=0$ & $x^2+y^2-4x-2y-6=0$ is $\frac{\alpha \sqrt{106}}{\beta }$ where $\alpha$ and $\beta$ are coprime then

• A. $\alpha -\beta =3$
• B. $\alpha -\beta =2$
• C. $\alpha +\beta =7$
• D. $\alpha +\beta =5$

Answer 1

The correct option is C $\alpha +\beta =7$

Let $C_1:x^2+y^2+2x+6y=0$ and $C_2:x^2+y^2-4x-2y-6=0$

Consider equation of first circle.

$\therefore x^2+y^2+2x+6y=0$

$\therefore (x^2+2x)+(y^2+6y)=0$

$\therefore (x^2+2x+1)-1+(y^2+6y+9)-9=0$

$\therefore {(x+1)}^2+{(y+3)}^2=10$

$\therefore {(x-(-1))}^2+{(y-(-3))}^2={\left(\sqrt{10}\right)}^2$

Thus, coordinates of center of first circle are, $C_1(-1,-3)$ and radius is $r_1=\sqrt{10}$

Now, consider equation of second circle.

$x^2+y^2-4x-2y-6=0$

$\therefore (x^2-4x)+(y^2-2y)-6=0$

$\therefore (x^2-4x+4)-4+(y^2-2y+1)-1-6=0$

$\therefore {(x-2)}^2+{(y-1)}^2=11$

$\therefore {(x-2)}^2+{(y-1)}^2={\left(\sqrt{11}\right)}^2$

Thus, coordinates of center of second circle are, $C_2(2,1)$ and radius is $r_2=\sqrt{11}$

Now, Let A and B are end points of common chord to both the circles. As the chord is common, we can write as,

$S_2=S_1$

$\therefore S_2-S_1=0$

$\therefore C_2-C_1=0$

$\therefore (x^2+y^2-4x-2y-6)-(x^2+y^2+2x+6y)=0$

$\therefore x^2+y^2-4x-2y-6-x^2-y^2-2x-6y=0$

$\therefore -6x-8y-6=0$

$\therefore 3x+4y+3=0$ (Dividing by -2 on both sides)

This is equation of common chord.

Now, Let us draw perpendicular from point $C_2$ on a common chord. Let M be foot of perpendicular.

Now, Distance between point $C_2$ and common chord is calculated by following formula-

$d\left(C_{2}M\right)=\frac{|Ah+Bk+C|}{\sqrt{A^2+B^2}}$

where, A = Coefficient of x in equation of common chord

B = Coefficient of y in equation of common chord

C = Constant term in equation of common chord

Here, $A=3$, $B=4$ and $C=3$

$\therefore d\left(C_{2}M\right)=\frac{|(3\times 2)+(4\times 1)+3|}{\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}}$

$\therefore d\left(C_{2}M\right)=\frac{6+4+3}{\sqrt{25}}$

$\therefore d\left(C_{2}M\right)=\frac{13}{5}$

Now, consider right angled triangle $C_{2}AM$. By Pythagoras theorem,

${\left(C_{2}A\right)}^2={\left(C_{2}M\right)}^2+{\left(MA\right)}^2$

But, $d\left(C_{2}A\right)=r_2$

$\therefore {\left(r_2\right)}^2={\left(C_{2}M\right)}^2+{\left(MA\right)}^2$

$\therefore {\left(\sqrt{11}\right)}^2={\left(\frac{13}{5}\right)}^2+{\left(MA\right)}^2$

$\therefore 11=\frac{169}{25}+{\left(MA\right)}^2$

$\therefore {\left(MA\right)}^2=11-\frac{169}{25}$

$\therefore {\left(MA\right)}^2=\frac{275-169}{25}$

$\therefore {\left(MA\right)}^2=\frac{106}{25}$

Taking square roots on both the sides, we get,

$\therefore MA=\frac{\sqrt{106}}{5}$

Now, we know that perpendicular drawn from center of circle on chord bisects the chord.

$\therefore AB=2\times MA$

$\therefore d\left(AB\right)=2\times \frac{\sqrt{106}}{5}$

Compare this equation with the one given in equation i.e.$\frac{\alpha \sqrt{106}}{\beta }$, we can write,

$\alpha =2$ and $\beta =5$

$\therefore \alpha +\beta =2+5=7$

Thus, Answer is option (C)