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sim0nz12345 · Feb 19, 2007, 10:43 PM

Hi
Can you please answer this question and show me the solution, cheers.

Show that the line 3x-4y+15=0 is a tangent to the circle x^2+y^2=9

Thanks

Capuchin · Feb 19, 2007, 11:27 PM

You need to prove that your two curves cross at only one point, that defines a tangent.

How would you do this, do you think?

sim0nz12345 · Feb 19, 2007, 11:38 PM

I'm not sure. I don't really understand the concept of tangents and the unit circle.
I just need the solution to it then it will get easier.

Capuchin · Feb 20, 2007, 12:36 AM

I just told you what a tangent is, it's a line that crosses at only one point (not 2 or 0 points).

You don't have a unit circle, you have a circle of radius 3, so you don't need to worry about that.

asterisk_man · Feb 20, 2007, 07:54 AM

1: find the point(s) of intersection between the two equations given
if step 1 finds more or less than 1 point of intersection then the line isn't tangent to the circle and you can quit here
2: find the slope of the circle at the point of intersection and the slope of the line
if the two slopes are equal then you've got a tangent line, otherwise you do not.

If you need help completing those steps please ask here.

sim0nz12345 · Feb 20, 2007, 01:55 PM

I not entirely sure how to find the intersection of the first step.
I know simultaneous equations is invloved but I don't know what to do with the squared terms.

asterisk_man · Feb 20, 2007, 02:07 PM

try solving the linear equation for one variable, substitute into the circle equation, solve for the remaining variable, substitute that value into the linear equation solved for 1 var to find the other variable.

sim0nz12345 · Feb 20, 2007, 10:28 PM

Ok thanks
I eventually found the answer to be a perfect square which means that there can only be one point of intersection, in a tangent.
x=-9/5
y=12/5
What do I then do for the next step

galactus · Feb 21, 2007, 05:18 AM

Plug your answers into the circle equation. Do you get 9? Plug them into the linear equation. Do you get 0?

sim0nz12345 · Feb 21, 2007, 01:53 PM

Yes it does.
Then what is the next step?

galactus · Feb 21, 2007, 02:09 PM

Next step? That's it.

asterisk_man · Feb 21, 2007, 02:19 PM

OK. I guess I agree. I was wrong. You don't need to compare the slopes. The only way that a line can have a single point of intersection with a circle is if it is tangent to the circle at that point. I agree with galactus. Your math is done. You just need to explain what I just said.