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# How to find chord length when a line intersects a circle?

Question:

A line, x - 7y + 25 = 0, intersects a circle, x2 + y2 = 25, at points E and F. What is the chord length EF?

Solution:

The general line equation is: Ax + By + C = 0

The standard circle equation is: (x - xo)2 + (y - yo)2 = r2, in which xo is the x-coordinate of the center of the circle. yo is the y-coordinate of the center of the circle. r is the radius of the circle.

A line, Ax + By + C = 0, intersects a circle, (x - xo)2 + (y - yo)2 = r2, at points E and F. So, EF is a chord. We need to find the chord length EF. Now we draw a line passes the point O (point O is the center of the circle) and perpendicular to chord EF. This line intersects EF at point G. G is the perpendicular point. Because OG is passing the center of a circle, so OG is a part of a diameter of the circle.

Because OG is perpendicular to the chord EF, so OG bisects the chord EF. (The diameter perpendicular to a chord bisects the chord)

Connect OE and OE = r

Let OG = d, which is the distance from the center of the circle to the chord EF.

EG = GF = (1/2)EF
In right triangle EGO,
(EF/2)2 = r2 - d2
EF/2 = square root of (r2 - d2)
EF = 2 square root of (r2 - d2)

Compare standard circle equation with the given circle equation, we get xo = 0, yo = 0, and r = 5.

Compare the general line equation with the given line equation, we get A = 1, B = -7 and C = 25.

d is the distance from the center of the circle to the chord EF

d = |Axo + Byo + C|/square root of (A2 + B2)
= |1 × 0 + (-7) × 0 + 25|/square root of (12 + (-7)2)
= 25/square root of 50
= 25/square root of (25 × 2)
= 25/(5 × square root of 2)
= 5/square root of 2
d2 = 25/2
EF = 2 square root of (r2 - d2)
= 2 square root of (25 - 25/2)
= 2 square root of [25(1 - 1/2)]
= 2 square root of 25/2
= 2 × 5 ÷ square root of 2
= 5 square root of 2

The chord length EF is 5 square roots of 2. Watch the video for more details.