Question

If the circles $x^2 + y^2 - 2x - 2y - 7 = 0$ and $x^2 + y^2 + 4x + 2y + k = 0$ cut orthogenally, then the length of the common chord of the circles is

• A. 2
• B. $12/\sqrt{13}$
• C. 8
• D. 5

Answer 1

Answer: (B)

Given, circles are
$S_1 = x^2 + y^2 - 2 x - 2 y - 7 = 0$
$S_2 = x^2 + y^2 + 4x + 2y + k = 0$
Here, $ g_1 = - 1 , f_1 = - 1 . c_1 = - 7$
$g_2 = 2 , f_2 = 1 , c_2 = k$

Equation of common chord is $S_1 - S_2 = 0$
$ \Rightarrow \, x^2 + y^2 - 2 x - 2 y - 7 - x^2- y^2- 4 x - 2y - k = 0$
$ \Rightarrow \,- 6 x - 4 y - 7 — k = 0$
$ \Rightarrow \,6x + 4y + 7 + k = 0 \,\,\,\,\,\dots(i)$
Since, $S_1$ and $S_2$ cut orthogonally
$ \therefore \, 2 (g_1 g_2 + f_1 f_2) = c_1 + c_2$
$ \Rightarrow \, 2 (-2 - 1) = - 7 + k$
$ \Rightarrow \, -6 + 7 = k$
$ \Rightarrow \, k = 1$
Then, from Eqs. (i), we get
$6 x + 4 y + 8 = 0$
Now, length of the common chord
$r_{1} = \sqrt{1+1+7} = 3$
$ c_{1} =\left(1,1\right) $
Let $C_1M$ = perpendicular distance from centre $C_1(1,1)$ to the common chord
$6 x + 4 y + 8 = 0$
$ \Rightarrow C_{1}M = \frac{\left|6+4+8\right|}{\sqrt{6^{2} +4^{2}}} = \frac{\left|18\right|}{5^{2}} = \frac{18}{2\sqrt{13}} = \frac{9}{\sqrt{13}} $
Now, $ PQ = 2PM = 2 \sqrt{\left(C_{1}P\right)^{2} -\left(C_{1} M\right)^{2}}$
$ =2 \sqrt{9- \left(\frac{9}{\sqrt{13}}\right)^{2}}$
$ = 2\sqrt{9- \frac{81}{13}} $
$ =2 \sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}} $