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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]

 Apr 7, 2024
 #2
avatar+9664 
0

\(\begin{cases} x^2 + y^2 = 5\\ y= \dfrac{3x+2}4 \end{cases}\)

 

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.

 Apr 8, 2024

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