Question

The length of the $y$-intercept made by the circle $x^2+y^2-4x-6y-5=0$ is :

• A. $6$
• B. $\sqrt{14}$
• C. $2\sqrt{14}$
• D. $3$

Answer 1

Answer: (C)

$2\sqrt{14}$

To know the points on $y$ axis, where the circle above intersects, we substitute $x$ as $0$.

We thus get $y^2-6y-5=0$

$\Rightarrow y=\frac{6\pm \sqrt{36+20}}{2}=\frac{6\pm \sqrt{56}}{2}$

$\Rightarrow y=3\pm \sqrt{14}$

So, the circle intersects the $y$ axis at points $(0,3-\sqrt{14})$ and $(0,3+\sqrt{14}).$

Thus, $y$ intercept becomes $3+\sqrt{14}-3+\sqrt{14}=2\sqrt{14}$