# Thread: Equation of a Tangent to a Circle

1. ## Equation of a Tangent to a Circle

A Circle with center (-4,2) has a line tangent to it at (-2,6). Find the equation of the tangent.

I want to check if i got the right answer for this question.

I found the slope of the line of both points which came out to be $1/2$.

Since the the line is tangent the slope would be the reciprocal of $1/2$ which is -2.

Plug in the slope with the points of the tangent (-2,6)... and I got the equation $y=-2x+2$.

Does this seem right? It just looks like it's too easy...

2. That is great reasoning. I made a mistake in my original post the slope should be 2 just as jhevon has said.

3. Originally Posted by chrozer
A Circle with center (-4,2) has a line tangent to it at (-2,6). Find the equation of the tangent.

I want to check if i got the right answer for this question.

I found the slope of the line of both points which came out to be $1/2$.

Since the the line is tangent the slope would be the reciprocal of $1/2$ which is -2.

Plug in the slope with the points of the tangent (-2,6)... and I got the equation $y=-2x+2$.

Does this seem right? It just looks like it's too easy...
your reasoning is correct. but i think you made an error with the slope. the slope of the line connecting the two points is 2 i believe

...or maybe i need some sleep. in any case, double check your answer. when all said and done, your line should be y = -(1/2)x + 5

4. Originally Posted by Jhevon
your reasoning is correct. but i think you made an error with the slope. the slope of the line connecting the two points is 2 i believe

...or maybe i need some sleep. in any case, double check your answer. when all said and done, your line should be y = -(1/2)x + 5
Actually that equation, which your right about being the line connecting the two points, would be correct, but I have to find the equation of the line going through the points (-2,6) while at the same time being tangent to a circle with center (-4,2).

So everything is fine and I got the answer right?

5. Originally Posted by chrozer
Actually that equation, which your right about being the line connecting the two points, would be correct, but I have to find the equation of the line going through the points (-2,6) while at the same time being tangent to a circle with center (-4,2).

So everything is fine and I got the answer right?
yes, that's what i meant.

the slope of the line passing through (-4,2) and (-2,6) is $\frac {6 - 2}{-2 - (-4)} = \frac 42 = 2$. the tangent line is perpendicular to this line, so the slope of the tangent line is -1/2

so you want the line with slope -1/2 passing through (-2,6)

6. Originally Posted by Jhevon
yes, that's what i meant.

the slope of the line passing through (-4,2) and (-2,6) is $\frac {6 - 2}{-2 - (-4)} = \frac 42 = 2$. the tangent line is perpendicular to this line, so the slope of the tangent line is -1/2

so you want the line with slope -1/2 passing through (-2,6)
Oh yeah...you're right. I had my it $\frac {x - x}{y - y}$ instead of $\frac {y - y}{x - x}$

7. Originally Posted by chrozer
Oh yeah...you're right. I had my it $(x-x)/(y-y)$ instead of $(y-y)/(x-x)$
ok. glad we cleared that up. now you can get the solution

use $$\frac {y_2 - y_1}{x_2 - x_1}$$ to get $\frac {y_2 - y_1}{x_2 - x_1}$

8. Originally Posted by Jhevon
ok. glad we cleared that up. now you can get the solution

use $$\frac {y_2 - y_1}{x_2 - x_1}$$ to get $\frac {y_2 - y_1}{x_2 - x_1}$
Ok...thanks. Now time for me to go to sleep