Question

lf the length of the common chord of two circles $x^2+y^2+Sx+1=0$ and
$x^2+y^2+2\mu y-1=0$ is $2\sqrt{6}$ then $\mu =\dots \dots$.

• A. $\pm 2$
• B. $\pm 4$
• C. $\pm 8$
• D. $\pm 3$

Answer 1

The correct option is D $\pm 3$

$c_{1}E=\sqrt{r_1^2(\sqrt{6}{)}^2}$
$=\sqrt{r_1^2-6}$
And $E{C}_2=\sqrt{r_2^2-6}$
$c_1{c}_2=c_{1}E+E{c}_2=\sqrt{r_1^2-6}+\sqrt{r_2^2-6}$
$\sqrt{\frac{s^2}{4}+{\mu }^2}=\sqrt{\frac{s^2}{4}-7}+\sqrt{{\mu }^2-5}$
$\frac{s^2}{4}+{\mu }^2=\frac{5^2}{4}+{\mu }^2-12+2\sqrt{(\frac{s^2}{4}-7)({\mu }^2-5)}$
$\frac{12}{2}=\sqrt{(\frac{s^2}{4}-7)({\mu }^2-5)}$
$36=(\frac{s^2}{4}-7)({\mu }^2-5)$