Finding the Equations of Tangent Lines to a Circle

[mathdad]

Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5, 12) and (5, -12). Find the equations of the tangents.

I need the steps.

I know the circle has a radius of sqrt{169} or 13.

 

[MarkFL]

Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5, 12) and (5, -12). Find the equations of the tangents.

I need the steps.

I know the circle has a radius of sqrt{169} or 13.

Okay, let's take this step by step...first, where is the center of the circle?

 

[mathdad]

The center is (0, 0) and radius 13.

 

[MarkFL]

The center is (0, 0) and radius 13.

Okay...good! (Yes)

Now, how does the radius of a circle (drawn from the center to the point of tangency) relate the the line tangent to the circle? If you need to, draw a circle and then draw a radius, and the line tangent to the circle where the radius meets the circle.

 

[mathdad]

The distance from the center to the point of tangency is 13.

The radius from the center to the point of tangency is perpendicular.

 

[MarkFL]

The distance from the center to the point of tangency is 13.

The radius from the center to the point of tangency is perpendicular.

Yes, and so you need to find the slopes of the given radii, and then use the negative reciprocal of the slope of the radius as the slope of the tangent line, and the given point of tangency in the point-slope formula to find the equations of the tangent lines. :D

 

[mathdad]

Ok. I will work on this one tomorrow before going to work and post my work for both equations.

 

[mathdad]

We have:

radius = 13

points (5, 12), (5, -12) and (0, 0)

Let m_1 = slope 1

m_1 = (5 - 0)/(12 - 0)

m_1 = 5/12

y - 12 = 5/12(x - 5)

y = (5/12)x - (25/12) - 12

y = (-12/5)x - (169/12)

Is this one of the equations of the tangents?

Let m_2 = slope 2

m_2 = (-12-0)/(5-0)

m_2 = -12/5

y -(-12) = (-12/5)(x - 5)

y + 12 = (-12/5)x + 60/5

y = (5/12)x + 12 - 12

y = (5/12)x

Is this the second equation?

 

[MarkFL]

Let's work a more general version of the problem...that is, to find the equation of the line tangent to the circle:

$$(x-h)^2+(y-k)^2=r^2$$

at the point on the circle:

$$\left(x_1,y_1\right)$$

where:

$$y_1\ne k$$

When $y_1=k$, then we know the tangent line is simply $x=x_1$.

The slope of the line will be:

$$m=\frac{h-x_1}{y_1-k}$$

And so, using the point-slope formula, the tangent line is:

$$y=\frac{h-x_1}{y_1-k}\left(x-x_1\right)+y_1$$

Arranged in slope-intercept form, we have:

$$y=\frac{h-x_1}{y_1-k}x+\frac{x_1-h}{y_1-k}x_1+y_1$$

For the given problem, we have:

$$(h,k)=(0,0)$$

And so, our formula reduces to:

$$y=-\frac{x_1}{y_1}x+\frac{x_1}{y_1}x_1+y_1=-\frac{x_1}{y_1}x+\frac{x_1^2+y_1^2}{y_1}=-\frac{x_1}{y_1}x+\frac{r^2}{y_1}$$

The first point is:

$$\left(x_1,y_1\right)=(5,12)$$

And so the tangent line is:

$$y=-\frac{5}{12}x+\frac{169}{12}$$

The second point is:

$$\left(x_1,y_1\right)=(5,-12)$$

And so the tangent line is:

$$y=\frac{5}{12}x-\frac{169}{12}$$

Let's plot the circle and the two lines to verify:

[DESMOS=-34.23893796773567,51.449953593644956,-15.304682349252822,14.422600065796832]x^2+y^2=13^2;y=-\frac{5}{12}x+\frac{169}{12};y=\frac{5}{12}x-\frac{169}{12}[/DESMOS]

 

[mathdad]

I undetstand the question better now than when I first read it in the textbook.

 

[mathdad]

Let's work a more general version of the problem...that is, to find the equation of the line tangent to the circle:

$$(x-h)^2+(y-k)^2=r^2$$

at the point on the circle:

$$\left(x_1,y_1\right)$$

where:

$$y_1\ne k$$

When $y_1=k$, then we know the tangent line is simply $x=x_1$.

The slope of the line will be:

$$m=\frac{h-x_1}{y_1-k}$$

And so, using the point-slope formula, the tangent line is:

$$y=\frac{h-x_1}{y_1-k}\left(x-x_1\right)+y_1$$

Arranged in slope-intercept form, we have:

$$y=\frac{h-x_1}{y_1-k}x+\frac{x_1-h}{y_1-k}x_1+y_1$$

For the given problem, we have:

$$(h,k)=(0,0)$$

And so, our formula reduces to:

$$y=-\frac{x_1}{y_1}x+\frac{x_1}{y_1}x_1+y_1=-\frac{x_1}{y_1}x+\frac{x_1^2+y_1^2}{y_1}=-\frac{x_1}{y_1}x+\frac{r^2}{y_1}$$

The first point is:

$$\left(x_1,y_1\right)=(5,12)$$

And so the tangent line is:

$$y=-\frac{5}{12}x+\frac{169}{12}$$

The second point is:

$$\left(x_1,y_1\right)=(5,-12)$$

And so the tangent line is:

$$y=\frac{5}{12}x-\frac{169}{12}$$

Let's plot the circle and the two lines to verify:

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