Question 222182
write the equation of the circle in radius-center form with the center at (2,1) and tangent to the line x=5.
<pre><font size = 4 color = "indigo"><b>
The equation of a circle in radius-center form is

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

where the center is the point {{{"(h,k)"}}} and
the radius is {{{r}}}.

So the equation, all but r, is:

{{{(x-2)^2+(y-1)^2=r^2}}}

So to find the radius {{{r}}} let's plot the center
{{{"(2,1)"}}}

{{{drawing(200,200,-2,6,-3,5,graph(200,200,-2,6,-3,5),
line(2+.1,1,2-.1,1), line(2,1+.1,2,1-.1),locate(2,1,"(2,1)"), 
line(2+.1,1+.1,2-.1,1-.1), line(2+.1,1-.1,2-.1,1+.1)
)}}} 


Now let's draw the line {{{x=5}}}, which is a vertical
line through 5 on the x-axis:

{{{drawing(200,200,-2,6,-3,5,graph(200,200,-2,6,-3,5),
line(2+.1,1,2-.1,1), line(2,1+.1,2,1-.1),locate(2,1,"(2,1)"), 
line(2+.1,1+.1,2-.1,1-.1), line(2+.1,1-.1,2-.1,1+.1),
line(5,-3,5,7)

)}}} 





Now let's sketch in a circle tangent to that line:

{{{drawing(200,200,-2,6,-3,5,graph(200,200,-2,6,-3,5),
line(2+.1,1,2-.1,1), line(2,1+.1,2,1-.1),locate(2,1,"(2,1)"), 
line(2+.1,1+.1,2-.1,1-.1), line(2+.1,1-.1,2-.1,1+.1),
line(5,-3,5,7),circle(2,1,3)

)}}} 

Draw a radius from the center (2,1) over to the line.

{{{drawing(200,200,-2,6,-3,5,graph(200,200,-2,6,-3,5),
line(2+.1,1,2-.1,1), line(2,1+.1,2,1-.1),locate(2,1,"(2,1)"), 
line(2+.1,1+.1,2-.1,1-.1), line(2+.1,1-.1,2-.1,1+.1),
line(5,-3,5,7),circle(2,1,3), line(2,1,5,1)

)}}}

Using the x-axis as a measuring stick, you can see that
that radius is 3 units long, so

{{{r=3}}}

and the final answer is

{{{(x-2)^2+(y-1)^2=3^2}}}  or

{{{(x-2)^2+(y-1)^2=9}}}

Edwin</pre>