Intersection of tangent line with circle

jonnburton

Junior Member
Joined
Dec 16, 2012
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155
I wondered if anyone could show me where I am going wrong with a question I am attempting:

"Show that the lengths of the tangents from the point (h,k) to the circle [math]x^2 + y^2 +2fx + 2gy + c + 0[/math] are [math]\sqrt{h^2+k^2+2fh+2gk+c}[/math]"

Completing the square to obtain the co-ordinates of the circle's centre:

[math](x+f)^2 -f^2 + (y+g)^2 -g^2 +c =0[/math][math](x+f)^2+(y+g)^2 = f^2+g^2-c[/math]
Centre point = (-f,-g)
Radius^2 = [math]f^2+g^2-c[/math]
Distance from (h,k) to circle's centre:
[math]\sqrt{(h-(-f))^2 + (k-(-g))^2}[/math][math]\sqrt{(h+f)^2 + (k+g)^2}[/math]
Using Pythagoras' theorem, the length of the tangents is:

[math](\sqrt{f^2+g^2-c})^2 + (\sqrt{(h+f)^2+(k+g)^2})^2[/math][math]=\sqrt{f^2+g^2-c+h^2+2hf+f^2+k^2+2gk+g^2}[/math][math]=\sqrt{2f^2+2g^2+f^2+k^+h^2+2hf+2gk-c}[/math], which is different from the correct answer

I'd be very grateful if anyone can show me what I am doing wrong here
 
A line from (h,k) (which is the center of the circle) to a point on the circle is called the radius of the circle and it's length is r. You found out that r^2 = f^2 + g^2 -c, so r is sqrt(f^2 + g^2 -c). Now I must admit that I not sure what you mean by the lengths of the tangents from the point (h,k) to the circle.
 
A line from (h,k) (which is the center of the circle) to a point on the circle is called the radius of the circle and it's length is r. You found out that r^2 = f^2 + g^2 -c, so r is sqrt(f^2 + g^2 -c). Now I must admit that I not sure what you mean by the lengths of the tangents from the point (h,k) to the circle.

Who said (h,k) is the center of the circle? Those letters are often used for a center, but not in this problem.

The center is (-f, -g). Point (h, k) is, presumably, outside the circle.
 
Yes, although I copied the question ad verbatim, I should have clarified (h,k) lies outside the circle and its the lengths of the tangents from that point to where they meet the circumference that's required. My mistake, which blamocur enabled me to see was that, having found the radius and distance from (h,k) to the centre, I somehow thought the tangent was the hypotenuse of the ensuing triangle, when in fact the hypotenuse is the line from (h,k) to the centre
 
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