Question 1050845: Find the equation of the circle of radius √29 tangent to the line 2x-5y+16=0 and passing through (9,-3)
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443)   (Show Source):
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! Here is your line, and your point  :
 The center of the circle is a point  at a distance  from P.
Once we find  and  , the equation of the circle will simply be
 .
The and your teacher has in mind are  , so the equation your teacher wants is
There is another point,  , that could also be the center of the circle, but your teacher does not know or does not care.
Otherwise the problem would say to find "the equation of the circles", ot "of one of the circles."
Anyway, the coordinates of that other circle center point are not such nice integer numbers, but I will tell you what the alternate solution is , anyway.
Since P is so much farther than 5.4 from your line the center of your circle must be below the line,
between the point and the line, and at a distance from the line.
All the points at a distance from from your line are on another line,
a line parallel to and below  .
The distance between the two lines is ,
but that is measured along a perpendicular to the lines,
not along a vertical line.
For  , gives you  as the x-intercept.
We could also calculate the y-intercept:
For  ,  --->  --->  .
From the coefficients in you see that
as x increases by  , y increases by  ,
meaning that has a slope of  ,
and so does any parallel line.
Below is a sketch showing the two parallel lines, and the distance between them.
To make it simpler, I did not draw the axes, but I drew some right triangles.
The sides of those triangles that look horizontal and vertical are really parallel to the axes.
The triangle on the right is drawn so that the hypotenuse measures from P to C the green line.
The triangle on the left could have been drawn anywhere,
as the legs are constructed to measure and ,
for a slope of , but so that the length of the hypotenuse is
 .
 Because pairs of their sides are perpendicular,
the angles of the triangles at the left and in the middle are congruent,
and that makes the triangles similar.
Because they are similar, the ratio of hypotenuse to long leg is the same:
 ---->  .
That tells us that the green line containing the center of the circle is
translated down by  units.
So, with a slope of , and a y-interceptat
 , units below the  y-intercept of ,
the equation of the green line is
 .
That is the line where we find points C (and D).
How do we find those points?
They satisfy , because they are on the green line,
and  because they are at a distance of from .
Points C and D are one end of the hypotenuse of a right triangle, with horizontal and vertical legs.
We know there are infinity of right triangles with a hypotenuse.
One of them is shown on the left, with legs measuring  and  .
Could the triangle with C and P be congruent?
With point C units left and units up from ,
we would have  .
Is that a point on the green line?
Substituting  into , we get  ,
So,  is the center of one circle.
Any other circles? Yes, solutions of the system
 .
-->  -->  -->  -->  .
The solutions to  are and  .
Substituting into we get  to find .
Substituting into we get  to find  .
So, there are two circles, the one whose equation is shown at top, and the one with center at D, and the ugly equation
 :
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