Tangent of a Circle

**immunofort** · July 19th 2008, 09:54 PM

The question is

Find the equation of the tangent and normal to the circle x^2 + y^2 = 13 at the point t(-2,3)

I was thinking about this question and would the tangent be perpendicular to the straight line joining the (-2,3) and Origin?

**Chris L T521** · July 19th 2008, 10:05 PM

Originally Posted by immunofort

The question is

Find the equation of the tangent and normal to the circle x^2 + y^2 = 13 at the point t(-2,3)

I was thinking about this question and would the tangent be perpendicular to the straight line joining the (-2,3) and Origin?

I don't think so. From what I see, the lines don't pass through the origin.

To find the slope, we can use implicit differentiation:

$\frac{d}{dx}\bigg[x^2+y^2\bigg]=\frac{d}{dx}\bigg[13\bigg]$

$\implies 2x+2y\frac{dy}{dx}=0$

$\implies \frac{dy}{dx}=-\frac{x}{y}$

We can find the value of the slope of the tangent line at the point $(-2,3)$ :

$\bigg.\frac{dy}{dx}\bigg|_{x=-2, \ y=3}=-\frac{-2}{3}=\frac{2}{3}$

The slope of the normal line would be $-\frac{3}{2}$ .

Thus, the tangent line has the form $y=\frac{2}{3}(x-x_0)+y_0$

At the point $(-2,3)$ , the equation of our tangent line is $y=\frac{2}{3}(x+2)+3\implies \color{red}\boxed{y=\frac{2}{3}x+\frac{13}{3}}$

The normal line has the form $y=\frac{2}{3}(x-x_0)+y_0$

At the point $(-2,3)$ , the equation of our normal line is $y=-\frac{3}{2}(x+2)+3\implies \color{red}\boxed{y=-\frac{3}{2}x}$

I don't think so. From what I see, the lines don't pass through the origin.

I take that back. The tangent line is perpendicular to the normal line [and the normal line happens to pass through the origin].

I hope that this makes sense!

--Chris

**Henderson** · July 20th 2008, 12:16 AM

Originally Posted by immunofort

The question is

Find the equation of the tangent and normal to the circle x^2 + y^2 = 13 at the point t(-2,3)

I was thinking about this question and would the tangent be perpendicular to the straight line joining the (-2,3) and Origin?

Yes. It definitely will. Find your slope from the origin to (-2,3), find a perpendicular slope, and put that slope through (-2,3).

That is absolutely how you do this problem.

**gtbiyb** · July 20th 2008, 12:33 AM

Just a few points to add!

From your circle equation you should be able to recognise a few things. Firstly we know the center of the circle is the origin as it has not been translated by anything if it had it would look like (x - 2)^2 + (y - 5)^2 = 13 or similar.

Secondly any tangent of a circle will ALWAYS make a right angle with the radius at the point where it touches the circle. With that in mind the equation of the normal is simply the equation of the radius at that point.

So we have the coords for the center of the circle (0, 0) and we have the the coords for the point on the circumference (-2, 3) so with these we can find the gradient of the radius thus m = (3 - 0)/(-2 - 0) ie change in y divided by the change in x and we get m = - 3/2. To find the equation of the normal we need the gradient of the normal which we simply use the rule
m1 x m2 = -1. That is if two lines are at right angles then the product of there gradients = -1. so -3/2 x m2 = -1
m2 = -1/-3/2
m2 = 2/3
So the gradient of the normal is 2/3. so the eqation of the normal is worked out thus

2x + 4 = y – 3
3 3 NOTE - This should read 4/3 but i cant get the 3 under the 4!

2x + 4 = 3y – 9

2x + 13 = 3y

I would leave it as 3y – 2x = 13

Hope that adds to the above post!

**Chris L T521** · July 20th 2008, 12:39 AM

Originally Posted by gtbiyb

Just a few points to add!

From your circle equation you should be able to recognise a few things. Firstly we know the center of the circle is the origin as it has not been translated by anything if it had it would look like (x - 2)^2 + (y - 5)^2 = 13 or similar.

Secondly any tangent of a circle will ALWAYS make a right angle with the radius at the point where it touches the circle. With that in mind the equation of the normal is simply the equation of the radius at that point.

So we have the coords for the center of the circle (0, 0) and we have the the coords for the point on the circumference (-2, 3) so with these we can find the gradient of the radius thus m = (3 - 0)/(-2 - 0) ie change in y divided by the change in x and we get m = - 3/2. To find the equation of the normal we need the gradient of the normal which we simply use the rule
m1 x m2 = -1. That is if two lines are at right angles then the product of there gradients = -1. so -3/2 x m2 = -1
m2 = -1/-3/2
m2 = 2/3
So the gradient of the normal is 2/3. so the eqation of the normal is worked out thus

2x + 4 = y – 3
3 3 NOTE - This should read 4/3 but i cant get the 3 under the 4!

2x + 4 = 3y – 9

2x + 13 = 3y

I would leave it as 3y – 2x = 13

Hope that adds to the above post!

You're correct.

--Chris

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A few points!

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