# Tangents to circle

• Nov 25th 2008, 04:09 AM
Thomas154321
Tangents to circle
I can't work out how to answer this question, mental blockage or something!

Find the values of k such that $\displaystyle y=kx$ are the tangents to the circle $\displaystyle x^2+y^2-4x-6y+3=0$.

(Not exactly sure that's the circle I need but I think it should work as it doesn't cross any axes.)

Thanks!
• Nov 25th 2008, 04:49 AM
Quote:

Originally Posted by Thomas154321
I can't work out how to answer this question, mental blockage or something!

Find the values of k such that $\displaystyle y=kx$ are the tangents to the circle $\displaystyle x^2+y^2-4x-6y+3=0$.

(Not exactly sure that's the circle I need but I think it should work as it doesn't cross any axes.)

Thanks!

(Hi)Its a circle
$\displaystyle (x-2)^2+(y-3)^2={\sqrt{10}}^2$
Its a circle with centre at (2,3) and radius $\displaystyle \sqrt{10}$
(Wait):::::If you still need help:::
the line y= kx is tangent to the circle
$\displaystyle (1+k^2)x^2-(6k+4)x+3=0$
to be tangent there should be only one value of x so
$\displaystyle (6k+4)^2=12(1+k^2)$