# Line intersecting circle.xls

(4/1)

### Description

Purpose of calculation
Find intersection between a line and a circle with centre at origin.

Calculation Reference
Geometry for Enjoyment and Challenge, Rhoad

Calculation Validation
Check graphically in chart.
XLC addin used to verify cell formulas.

Calculation Reference

Geometry for Enjoyment and Challenge, Rhoad

Geometry

Coordinate geometry

To find the intersection between a line and a circle with center at the origin, we can use the general equation of a circle:

x^2 + y^2 = r^2

where (0,0) is the center of the circle, and r is the radius.

Let the equation of the line be y = mx + b, where m is the slope and b is the y-intercept.

Substituting y = mx + b into the equation of the circle, we get:

x^2 + (mx + b)^2 = r^2

Expanding the squares, we get:

x^2 + m^2x^2 + 2bmx + b^2 = r^2

Rearranging, we get:

(m^2 + 1) x^2 + 2bm x + (b^2 - r^2) = 0

This is a quadratic equation in x. Solving for x using the quadratic formula, we get:

x = [-bm ± sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)

Substituting this value of x into the equation of the line, we get the corresponding values of y:

y = mx + b

So the points of intersection between the line and the circle are:

(x1, y1) = ([-bm + sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1), m([-bm + sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)) + b)

(x2, y2) = ([-bm - sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1), m([-bm - sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)) + b)

These are the two points of intersection between the line and the circle.

### Calculation Preview

28 Apr 2023
File Size: 115.50 Kb