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charleyvarkey · Jun 14, 2009, 11:13 PM

consider the circle x2+y2+2x-2y+4=0 and P(2,3) be any point.Let T be the point of contact to the tangent from PT.

1) Find the length of the tangent PT?
2) Find the equation of the tangent PT?

Perito · Jun 15, 2009, 05:49 AM

$x^2+2x+y^2-2y+4=0$ (equation 1)

The general equation of a line is:

$(x-a)^2 + (y-b)^2 = r^2$

Therefore, equation 1 can be written:

$(x+1)^2 + (y-1)^2 = -2$ (equation 2).

This gives us the center of the circle (-1,1) and the radius $\sqrt{-2}$ ??

Are you sure that's the equation of a circle?

http://mathworld.wolfram.com/CircleTangentLine.html

http://en.wikipedia.org/wiki/Circle

Unknown008 · Jun 15, 2009, 10:58 AM

Originally Posted by Perito

$x^2+2x+y^2-2y+4=0$ (equation 1)

The general equation of a line is:

$(x-a)^2 + (y-b)^2 = r^2$

Therefore, equation 1 can be written:

$(x+1)^2 + (y-1)^2 = -2$ (equation 2).

This gives us the center of the circle (-1,1) and the radius $\sqrt{-2}$ ???

Are you sure that's the equation of a circle?

Circle Tangent Line -- from Wolfram MathWorld

Circle - Wikipedia, the free encyclopedia

Missed a word here Perito. You didn't have your coffee I supposed :p

** The general equation of a circle is:

$(x-a)^2 + (y-b)^2 = r^2$

galactus · Jun 15, 2009, 12:58 PM

Are you sure you do not have a typo in your circle equation. IUpon completing the square,

we end up with a negative radius.

Are you sure it is not $x^{2}+y^{2}+2x-2y-4=0$ ?

If so, this gives:

$(x^{2}+2x+1)+(y^{2}-2y+1)=6$

Factor:

$(x+1)^{2}+(y-1)^{2}=6$

It then has center (-1,1) and radius $\sqrt{6}$

The distance from the center of the circle to P(2,3) is

$\sqrt{(2-(-1))^{2}+(3-1)^{2}}=\sqrt{13}\approx 3.6$

Therefore, P(2,3) it is outside the circle.

The equation of the line from the center of the circle to P(2,3) is

$y=\frac{2}{3}x+\frac{5}{3}$

Now, is it safe to assume you know how to find the equation of a line given two points it passes through?

Assuming I understood correctly, I have attached a diagram you can work from.

Find the point where the line from the center to P(2,3) intersects the circle.

Then, you can use the distance formula to find that distance from P(2,3).

Then, find the equation of the tangent to the circle at that point, y=mx+b

Write back if you're still stuck.