hi, can someone show me in detail how to answer the question:
show that the lengths of the tangents from the point (h, k) to the circle $\displaystyle x^2 + y^2 + 2fx + 2gy + c = 0$ are $\displaystyle \sqrt {h^2 + k^2 + 2fh + 2gk + c}$
thankyou
hi, can someone show me in detail how to answer the question:
show that the lengths of the tangents from the point (h, k) to the circle $\displaystyle x^2 + y^2 + 2fx + 2gy + c = 0$ are $\displaystyle \sqrt {h^2 + k^2 + 2fh + 2gk + c}$
thankyou
The circle's standard equation is (x + f)^2 + (y + g)^2 = f^2 + g^2 - c ==> this is the circle with center (-f,-g) and radius sqrt(f^2 + g^2 - c).
Now use some geometry: draw a circle and a point exterior to it and draw:
== the line joining the point and the circle's center and
== the point and one of the two tangency point on the circle (this is just one of the two tangents determined by the point and the circle) and
== the radius to the tangency point chosen.
You get a straight angle triangle whose straight angle is between the tangent and the radius (this is a basic theorem in geometry) ==> the distance you're looking for is the length of the catetus formed by the point and the tangency point ==> by Pythagoras theorem, this distance squared equals the square distance between the point and the circle's center minus the square of the radius.
Well, now just do the maths.
Tonio
Hello markThe centre of the circle is at $\displaystyle (-f,-g)$ and its radius, $\displaystyle r = \sqrt{f^2+g^2 -c}$. The distance of $\displaystyle (-f,-g)$ from $\displaystyle (h,k)$ is $\displaystyle \sqrt{(h+f)^2+(k+g)^2}$.
The tangent is perpendicular to the radius through the point of contact. So using Pythagoras on the triangle formed from the centre, the point $\displaystyle (h,k)$ and the point of contact, the length of the tangent, $\displaystyle l$, is given by:
$\displaystyle l^2 = (h+f)^2+(k+g)^2 -r^2$
$\displaystyle =(h+f)^2+(k+g)^2 -(f^2+g^2-c)$
$\displaystyle \Rightarrow l= \sqrt{h^2 +k^2+2fh + 2gk +c}$
Grandad