# Circle Exercise Question

## Homework Statement

Equation: x^2+y^2-6x-2y+8=0 Find the center and the radius.
(Help) : Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)

## The Attempt at a Solution

I found the center and radius and i believe the values are : C (3,1) and R =root(2) but can you explain me the second one

Sorry for my english.

I hope you'll understand it.

Thank you.

.Scott
Homework Helper
(x-3)^2 + (y-1)^2 -2 = x^2-6x+9+y^2-2y+1 -2 = x^2 + y^2 - 6x - 2y +10 - 2

Since your tangent line crosses through (0,0), it will be of the form x=my, where m is the slope. It will be perpendicular to your circle, so the formula for a line crossing through the tangent point and the center of the circle will be (x-3) = (-1/m)(y-1).

Now you need to find a slope "m" that will create an intersection between those equations that is on the circle: (x^2 + y^2 - 6x - 2y +8 = 0).

Also:
If you draw a picture, you will notice that there are two solutions. In both cases, a right triangle can be formed between (0,0), (3,1), and the tangent point. In both cases, the length of the hypotenuse will be sqrt(3^2+1^2) and the length of the leg that extends from the center of the circle will be sqrt(2). So the distance of the tangent points to the origin (0,0) can be easily computed.
So both tangent points will be on a circle centered at (0,0) with that radius. I'll let you set up that equation.

cristo
Staff Emeritus

## The Attempt at a Solution

I found the center and radius and i believe the values are : C (3,1) and R =root(2) but can you explain me the second one

How did you find the center and radius?

.Scott
Homework Helper
How did you find the center and radius?
I'm assuming that you only want an answer from the OP on this.

cristo
Staff Emeritus