how do you find the tangent line of a circle?

Question

how do you find the equation of the tangent line of the circle with radius 2 at origin (0,0)? The tangent line passes through the point (8,0) and the tangent line has to be in the first quadrant. Help me please.

Answer 1

You can try using the equation of the circle (x² + y² = 4), solving for y and finding the derivative to get the slope, etc. But another method is to realize that the tangent line of a circle is going to be perpendicuar to the radius at that same point. This means that the origin, the tangent point, and (8,0) form a right triangle. The hypotenuse is 8, the left leg is 2, and the other leg is the length of the tangent line between (8,0) and where it touches the circle. The length of this would be, using Pythagoras: √(8² - 2²) = √60.

Since x² + y² = 4 is the equation for the circle, any point on the circle is going to be in the form (x, ±√(4 - x²)). We're only concerned with the first quadrant right now, so we can drop the ± sign. Let's say the tangent point is at (x1, y1), or (x1, √(4 - x1²)). The distance between this and (8,0) must be √60. So use the distance formula:
(√60)² = (8 - x1)² + (0 - √(4 - x1²))²
60 = (8 - x1)² + (4 - x1²)
60 = (64 - 16x1 + x1²) + (4 - x1²)
-8 = -16x1 + x1² - x1²
-8 = -16x1
x1 = 1/2.
Therefore y1 = √(4 - (1/2)²) = √(16/4 - (1/4)) = √(15)/2

You've got 2 points of the line now, so you can write the equation for the line using point-slope form:

y - y2 = m(x - x2)
y - 0 = [(√(15)/2 - 0) / (1/2 - 8)](x - 8)
y = [(√(15)/2) / (-15/2)](x - 8)
y = [( -√(15)/15 )](x - 8)
y = [-√(15)/15]x + 8√(15)/15

Answer 2

let the tangent meet the circle at P (a,b)

equation of the circle with center as origin is x^2 +y^2 = r^2, r is the radius. Hence equation of this circle is x^2 +y^2 = 4.

since p is on the circle, it has to satisfy the equation of the circle. hence, a^2 + b^2 = 4.

the radius of a circle is always perpendicular to the tangent. since the radius passes through (0,0) and (a,b), slope of radius is b/a.
hence slope of tangent is: -a/b

we also know that the tangent passes through (8,0) and (a,b). hence slope of tangent is b/(a-8)

hence:
b/(a-8) = -a/b
b^2 = -(a^2) + 8a
b^2 + a^2 = 8a

we already know a^2 + b^2 = 4. hence:
8a = 4 hence a = 1/2

substituting this in a^2 + b^2 = 4 we have, b = root (15)/2 = 0.97

hence we have P (1/2,0.97)
slope of the line is -a/b = - 0.515
hence equation is y = -0.515 (x - 8)
y + 0.515x = 4.12

Answer 3

with the intention to locate the equation of a line, you want the slope and a level that you comprehend is on the line. The slope is immediately ahead: a tangent to a circle is perpendicular to the radius on the point the position the line will be tangent to the circle. In different words, the radius of your circle starts at (0,0) and is going to (3,4). This slope is an similar as upward push/run = (y2-y1)/(x2-x1) = (4-0)/(3-0) = 4/3. A line it really is perpendicular to this radius line would have a slope it really is the unfavorable reciprocal of the slope you comprehend. In different words, the line we decide would have a slope of -3/4. the point all of us comprehend is on the line is (3,4). So the widely used equation of the line is: (y-y0) = m(x-x0), the position (x0, y0) is the point all of us comprehend is on the line, or (3,4) for that reason m is the slope, -3/4 for that reason x,y stay as variables (y - 4) = (-3/4)(x - 3) y - 4 = (-3/4)x + 9/4 y = (-3/4)x + 25/4

Answer 4

Hint: The tangent line is perpendicular to the radius
at any point.