+39 A line and a circle intersect at A and B as shown below. Find the distance between A and B.
Asymptote code below!
[asy]
usepackage("amsmath");
usepackage("amssymb");
unitsize(2 cm);
pair A, B;
A = dir(210);
B = dir(50);
draw(Circle((0,0),1));
draw(interp(A,B,-0.2)--interp(A,B,1.2));
dot("$A$", A, W);
dot("$B$", B, N);
label("$x^2 + y^2 = 5$", dir(135), NW);
label("$y = \dfrac{3x + 2}{4}$", interp(A,B,1.2), E);
[/asy]
+9583 \(\begin{cases} x^2 + y^2 = 5\\ y= \dfrac{3x+2}4 \end{cases}\)
Substituting, we have
\(x^2 + \left(\dfrac{3x + 2}4\right)^2 = 5\\ \dfrac{25}{16}x^2 + \dfrac34x - \dfrac{19}4 = 0\\ 25x^2 + 12x - 76 = 0\\ (x + 2)(25x - 38) = 0\\ x = -2 \text{ or }x = \dfrac{38}{25}\)
When x = -2, y = (3(-2) + 2)/4 = -1.
When x = 38/25, y = (3(38/25) + 2)/4 = 41/25.
The required distance is:
\(\sqrt{\left(-2 - \dfrac{38}{25}\right)^2 +\left(-1 - \dfrac{41}{25}\right)^2 } = \dfrac{22}5\) units.
