There will be two solutions because the circle could be on either
side of the line.
The equation of a circle is:
(x-h)² + (y-k)² = r² with center (h,k) and radius r.
The radius is the distance from (h,k) to (-1,8).
We use the distance formula:
Square both sides:
The radius is also the perpendicular distance from
(h,k) to the line 3x+y-5=0
We use the formula:
The perpendicular distance from the point (x1,y1)
to the line Ax+By+C=0 is d =
=
=
=
=
=
=
=
Substitute in
There is a lot of work here! I'm not going to type the
steps. But you square those out and use the quadratic
formula.
Use the + and solve for h and get
then
Then substitute that in:
So the center of one circle is
(h,k) =
To find the equation we substitute for h,k, and r in
Using the - sign, and doing the same thing:
(h,k) =
You can get the equation of that circle the same way.
.
Find the equation of a circle that has radius length {sqrt(30)} and is tangent to the line 3x+y-5=0 at the point (-1,8).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I think that more short (and more straightforward) solution is possible.
The idea is to draw (to write the equation of) the straight line perpendicular to the given line through the given point (-1,8),
and then to take its intersection with the circle of the radius centered at the given point (-1,8).
Plot y = 5-3x (the given line, red), y = (the perpendicular at (-1,8), green)
and the circle centers (intersection points of the green straight line with the arcs, blue and purple arcs)
1. The straight line perpendicular to the given line 3x+y-5=0 at the point (-1,8) is
-(x-(-1)) + 3(y-8) = 0, or (which is the same) -x - 1 + 3y - 24 = 0 or (which is equivalent) x = 3y-25.
2. The circle of the radius centered at (-1,8) is
= .
3. The centers of two circles we are searching for, are the intersection points, i.e. the solutions of the system
x = 3y-25, (1)
= . (2)
To solve the system, substitute (1) into (2), replacing x. You will get
= 30,
9y^2 - 144y + 576 + y^2 - 16y + 64 = 30,
10y^2 - 160y + 610 = 0,
= = = .
Thus the centers are
a) = , = 3y-25 = , and
b) = , = 3y-25 = .
Having the centers and the radius r = , everybody can write the standard equations for the two circles.