# Tangents to circle

• November 25th 2008, 05:09 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 $y=kx$ are the tangents to the circle $x^2+y^2-4x-6y+3=0$.

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

Thanks!
• November 25th 2008, 05:49 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 $y=kx$ are the tangents to the circle $x^2+y^2-4x-6y+3=0$.

(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
$
(x-2)^2+(y-3)^2={\sqrt{10}}^2
$

Its a circle with centre at (2,3) and radius $\sqrt{10}$
(Wait):::::If you still need help:::
the line y= kx is tangent to the circle
$
(1+k^2)x^2-(6k+4)x+3=0
$

to be tangent there should be only one value of x so
$
(6k+4)^2=12(1+k^2)
$