Question

3. Show that a circle with equation (x−2)^2+(y+3)^2=160 has two tangent lines with equations

y+15=1/3(x−6) and y−9=1/3(x+2).

Answer 1

see below

Explanation:

A graphing calculator shows each of the lines intersect the circle in exactly one place, so is tangent.

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Each of the given lines is written in point-slope form. Each has a slope of 1/3. We will call the point in its equation the "reference point" of each line. For the first line, it is (6, -15); for the second line it is (-2, 9).

The center of the circle is (2, -3), so the slope from the center to the reference point of each line is ...

line 1: (-15 -(-3))/(6 -2) = -12/4 = -3

line 2: (9 -(-3))/(-2-2) = 12/-4 = -3

These slopes are the negative reciprocal of the slope of each of the lines, so a segment from the circle center to the reference point of each line is perpendicular to the line.

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The reference point of line 1 (6, -15) is on the circle:

(6 -2)^2 +(-15 +3)^2 = 16 +144 = 160

The reference point of line 2 (-2, 9) is on the circle:

(-2-2)^2 +(9 +3)^2 = 16 +144 = 160

So, the reference point of each line is on the circle and that point is the endpoint of a segment from the circle center that is perpendicular to the line. Hence each of the lines is a tangent at its reference point.