# Questions about finding values of a tangent

1. Aug 24, 2011

### J-Girl

Hi there:) so i have gotten this question in my homework, and its not that i dont understand how to find a tangent to a curve, i just dont understand this question!!!:(:( so here it is:
"The line x+y-k=0 is a tangent to the circle x^2+y^2-2x+4y-72=0. Find the value(s) of k.
Does this mean i am to solve the equation as a quadratic, and obtain the values of x to find the values of k?
I have tried to solve it as y=sqrt(-x^2+2x-4y+72),but the graph turned out totally different from the first one.
could anybody hely me? would really appreciate it!!!:)

2. Aug 24, 2011

### brunnels

You don't need to do any calculus to solve this problem.
Do a little algebra to show that the circle is centered at (1,-2) and has a radius of sqrt(77).
You know that the slope of the line is dy/dx = -1, and you should know intuitively what points on the circle should have that slope.
Then all you need to do is calculate where those points are and you can solve for the necessary k to have the line pass through the points.

3. Aug 24, 2011

### HallsofIvy

Staff Emeritus
Although brunnels says, you don't really need to use Calculus. (Although I see that he then refers to the derivative!) A line either misses a circle completely, or crosses it in two points, or is tangent to it. Of course, if the line is either tangent to the circle or crosses it in two points, the (x, y) coordinates at a point of intersection must satisfy both equations:
y= -x+ k and $x^2+ y^2- 2x+ 4y= 72$

Replace each y in the equation of the circle by -x+ k and you have a quadratic equation for equation in x. Now, under what conditions does that equation have 0, 1, or 2 solutions? What must k equal so that the equation has only one solution?