Circle Geometry at it's hardest

Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo

Reply 1

ian.slater

12

Original post by van.domo

Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo

There may be a typo in your question ... but the general idea would be to get the equation of the circle into the usual form to get the centre and radius.

Now work out the distance from (h,k) to the centre. Draw a diagram showing the tangent and radius at that point ....

Reply 2

ztibor

10

Original post by van.domo

Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo

Le t point P(h,k)
The line passing through the midpoint of the circle intersects the circle
at point A and B, then PA*PB=e^2
where e is the length of the tangent to the circle from P.
This is the theorem of power of P relating to (or on) the circle
The circle: