Question

Circle with Center (-1,2) and radius 5 goes through the line x+4y=0 in the dots A and B. Calculate the length of AB

Answer 1

From the line equation, we can deduce

$x=-4y$

This implies that any point on the line can be written as

$(-4y,y),\quad y \in \mathbb{R}$

So, given any two points

$A=(-4y_1,y_1),\quad B=(-4y_2,y_2)$

their distance will be

$AB=\sqrt{(-4y_1+4y_2)^2+(y_1-y_2)^2}=\sqrt{17(y_1-y_2)^2}=\sqrt{17}|y_1-y_2|$

So, we can compute the distance between two points on the line just by knowing their y coordinates!

Now, the equation of a circle with center $(-1,2)$ and radius 5 is

$(x+1)^2+(y-2)^2=5^2$

which we can reform as

$x^2 + y^2 + 2 x - 4 y - 20 = 0$

The points of intersection between the circle and the line are given by the system between the two equations:

$\begin{cases}x^2 + y^2 + 2 x - 4 y - 20 = 0\\x+4y=0\end{cases}$

From the second equation we can deduce $x=-4y$. Plugging this value in the first equation yields

$(-4y)^2 + y^2 + 2 (-4y) - 4 y - 20 = 0 \iff 17y^2-12y-20=0$

Solving this equation for $y$ yields

$y=\dfrac{12\pm\sqrt{144+1360}}{34}=\dfrac{12\pm\sqrt{1504}}{34}=\dfrac{12\pm\sqrt{16\cdot 94}}{34}=\dfrac{12\pm4\sqrt{94}}{34}=\dfrac{6\pm2\sqrt{94}}{17}$

Now we have the two $y$ values

$y_1=\dfrac{6+2\sqrt{94}}{17},\quad y_2=\dfrac{6-2\sqrt{94}}{17}$

Which implies that the distance length of AB is

$AB=\sqrt{17}|y_1-y_2|=\sqrt{17}\cdot\dfrac{4\sqrt{94}}{17}=4\sqrt{\dfrac{94}{17}}$

Answer 2

Answer:

$4\sqrt{1598}$

Step-by-step explanation:

distance center (–1, 2) to line x + 4y = 0 is

$d=\frac{-1+4(2)}{\sqrt{1^2+4^2}}=\frac{7}{17}\sqrt{17}$

by using Pythagorean's Theorem, we get

$AB=2\sqrt{r^2-d^2}\\=2\sqrt{25-\frac{49}{17}}\\=4\sqrt{1598}$