Question 1107104
A circle with centre (2,-3) has a tangent of 24x+7y-12=0. Find the equation of the circle.
<pre>
Here's the graph:

{{{drawing(400,400,-2,4,-4,2,
circle(2,-3,.05),green(line(1.424,-3.168,2,-3)),
graph(400,400,-2,4,-4,2), line(-3,12,4,-12), circle(2,-3,.6) )}}}

You have the center.  You need the radius, which is the green line.

What you need is the distance formula from a point to a line:

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>)to the line

Ax+By+C=0 is

d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

Your line is 24x+7y-12 = 0

So you plug in A=24, B=7, C=-12 and (x<sub>1</sub>,y<sub>1</sub>) = (2,-3)

Substitute that into the formula for d.  That will be the radius r.

Then substitute (h,k) = (2,-3) and that value of r into

{{{(x-h)^2+(y-k)^2=r^2}}}

If you have any difficulty, tell me in the note form below and I'll
get back to you by email.

Edwin</pre>