Question

Equation of tangent to a circle from origin

Answer 1

For example:

The given circle is (x-5)^2 + (y-3)^2 = (3)^2  means the circle is touching x- axis. So, y=0 is one tangent.

Now equation of tangent to a circle of above form is
(y-3) =m(x-5) (+/-) r *sq-rt (m^2+1), where m is the slope of the tangent and r is the radius.

In this case the eqn will be (y-3) =m(x-5) + 3 *sq-rt (m^2+1) as the slope cannot be negative. This passes through (0,0). Put x=0 and y=0 and solve for m.

You will get two values of m, i.e 0 and 15/8.

Putting m=15/8 in (y-3) =m(x-5) + 3 *sq-rt (m^2+1) you'll get another equation of tangent 15x -8y =0.