SOLUTION: The length of the tangent from (1,1) to the circle 2x^2+2y^2+5x+3y+1=0 is

Question 866242: The length of the tangent from (1,1) to the circle 2x^2+2y^2+5x+3y+1=0 is
Answer by josgarithmetic(39627) (Show Source): You can put this solution on YOUR website!
Identify three points.
Point R on the line is (1,1).
Point P on the circle being the point of tangency is an unknown point P (x,y).
Point C, the center of the circle.

You want the equation for the circle in standard form so you (1) know point C and (2) can find y as a function of x for this circle. You need to know how to Complete The Square, in order to put into standard form.

Two more things you want to do:
Make expression for the slope of CP;and; Make expression for the slope of PR.
The product of the slopes must be , so setting up this equality will allow you to solve for the x value for these slopes AND especially for point P's x coordinate.

You should definitely draw a picture or graph sketch of all this.

If you can make sense of this description and process, then you will be able to solve the problem.
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