Question

write an equation of the tangent line to the cirlce (x-2)^2 + (y+3)^2 =9 at the point (3,2radical2 -3)

Answer 1

First find your derivative of the equation, this will be you slope.

Answer 2

Unless you haven't learned derivatives yet..

Answer 3

we didnt do derivative yet this is pre calc

Answer 4

First analyze your equation: \((x-2)^2 +(y+3)^2=9\)

Format for the equation of a circle: \((x-a)^2 +(y-b)^2 = r^2\)

You could rewrite your equaiton so that it takes this form : \(\implies (x-\color{red}2)^2 +(y-\color{red}{(-3)})^2 = \color{blue}{3^2}\)

So the center of your circle is at \((2~,~-3)\)

Answer 5

And it has a radius of 3,

Answer 6

yeah but what is the tangent line to the circle at point (3,2radical2 -3)

Answer 7

First of all, i'd like to clarify what your point is. Is it \((3~,~ 2\sqrt{2}-3)\) ?

Answer 8

yes

Answer 9

So the point given is going to be a point on the tangent line to circle, and we have found the center.

Answer 10

Sorry, I was a little confused at first.

Answer 11

The slope of the tangent line is going to be the negative reciprocal, since the radius and the tangent line are perpendicular to eachother.

Answer 12

So we find our slope: \(m = \dfrac{y_2-y_1}{x_2-x_1}\)
\[(x_1~,~y_1) = (2~,~-3)\]\[(x_2~,~y_2) = (3~,~ 2\sqrt{2}-3)\]\[m=\dfrac{2\sqrt{2}-3-(-3)}{3-2} = \frac{2\sqrt{2}-3+3}{1} = 2\sqrt{2}\]

Answer 13

Now \(m_{\perp} = -\dfrac{1}{m}\)

Therefore \(m_{\perp} = -\dfrac{1}{2\sqrt{2}}\)

Answer 14

Now we have \(\color{blue}{m_{\perp}} = -\dfrac{1}{2\sqrt{2}}\) and we have a point \((3~,~ \color{blue}{2\sqrt{2}-3)}\)

You can use this given information to write out the equation of a tangent line in slope intercept form: \(y= \color{blue}{m_{\perp}}x+\color{blue}{b}\)

Answer 15

Just to clarify that the slope of a tangent line is the negative reciprocal, since the tangent line is perpendicular to the circle at the point.