jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 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
"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