Question 706690
I have a solution (two solutions actually), but there has to be a better way to get to the solution, and the results are so complicated that I suspect a typo in the problem.
The midpoint of AB is (-3,3).
The slope of the line through A and B is
{{{(4-2)/(-1-(-5))=2/4=1/2}}}
The perpendicular bisector of AB is
{{{y-3=-2(x+3)}}} --> {{{y-3=-2x-6}}} --> {{{y=-2x-3}}}
The center of a circle passing through A and B has to be on that line, so its coordinates would be (h,k) with
{{{k=-2h-3}}}
The equation of the circle is
{{{x-h)^2+(y-k)^2=R^2}}} where {{{R}}} is the radius.
I can even substitute {{{k=-2h-3}}} and get as equation for our circle
{{{x-h)^2+(y+2h+3)^2=R^2}}}
Substituting the coordinates of B, I get an expression for {{{R^2}}}
{{{-1-h)^2+(4+2h+3)^2=R^2}}} --> {{{h^2+2h+1+(7+2h)^2=R^2}}} --> {{{h^2+2h+1+4h^2+28h+49=R^2}}} --> {{{5h^2+30h+50=R^2}}}
Now I can write the equation for our circle as
{{{x-h)^2+(y+2h+3)^2=5h^2+30h+50}}}
Simple. I just have to find {{{h}}}.
I know the circle is tangent to the line
{{{x-5y=10}}} --> {{{x=5y+10}}}
so that line and the circle have just 1 intersection point.
Substituting {{{x=5y+10}}} into the equation for the circle,
{{{x-h)^2+(y+2h+3)^2=5h^2+30h+50}}}
I can find that intersection point and more.
{{{(5y+10-h)^2+(y+2h+3)^2=5h^2+30h+50}}}
{{{25y^2+100y-10hy+h^2+100-20h+y^2+6y+4h+4h^2+9+12h=5h^2+30h+50}}}
{{{26y^2+(106-6h)y+5h^2+109-8h=5h^2+30h+50}}}
{{{26y^2+(106-6h)y+59-38h=0}}}
If that quadratic equation must have just one solution, the discriminant must be zero, so
{{{(106-6h)^2-4*26*(59-38h)=0}}}
{{{11236-1272h+36h^2-6136+3952h=0}}}
{{{36h^2+2680h+5100=0}}}
The best I can do with that unwieldy equation is divide everything by 4 to get
{{{9h^2+670h+1275=0}}}
Applying the quadratic formula
{{{h = (-670 +- sqrt( 670^2-4*9*1275 ))/(2*9) }}}
{{{h = (-670 +- sqrt( 448900-45900 ))/18 }}}
{{{h = (-670 +- sqrt(403000))/18 }}}
That can be simplified to
{{{h = (-335 +- 5sqrt(4030))/9 }}}
Substituting into {{{k=-2h-3}}}, we get
{{{h = (-335 + 5sqrt(4030))/9 }}} with {{{k=(643-10sqrt(4030))/9}}} and
{{{h = (-335 - 5sqrt(4030))/9 }}} with {{{k=(643+10sqrt(4030))/9}}}
The approximate values give us points (-1.95,0.91) and (-72.49,141.98) for centers.
Exact values for {{{R^2}}} are
{{{R^2=(978475-15400sqrt(4030))/81}}} and {{{R^2=(978475+15400sqrt(4030))/81}}}
and approximate values are {{{R=3.235}}} and {{{R=155.4}}} corresponding to the centers above, respectively.
{{{drawing(300,300,-6,4,-3,7,
grid(0),
circle(-1.954,0.9086,3.2354),
line(-5,7,2,-7),
line(-5,2,-1,4),
line(10,0,-10,-4)
)}}} and {{{drawing(300,300,-300,100,-50,350,
grid(0),
circle(-72.49,141.98,155.4),
line(-200,397,100,-203),
line(-490,-100,510,100)
)}}}
The equation of the circles are
{{{(x-(-335 + 5sqrt(4030))/9 )^2+(y-(643-10sqrt(4030))/9)^2=(978475-15400sqrt(4030))/81}}} for the small circle, and
{{{(x+(335 + 5sqrt(4030))/9 )^2+(y-(643+10sqrt(4030))/9)^2=(978475+15400sqrt(4030))/81}}} for the large circle.
Asking for the equation of circle in general form is cruel.