SOLUTION: the circle passes through the points A(-5,2) and B(-1,4) and tangent to the line x-5y=10. 1. radius of the circle 2. equation of a circle in a center-radius form 3. equation of

Algebra ->  Circles -> SOLUTION: the circle passes through the points A(-5,2) and B(-1,4) and tangent to the line x-5y=10. 1. radius of the circle 2. equation of a circle in a center-radius form 3. equation of       Log On


   



Question 706690: the circle passes through the points A(-5,2) and B(-1,4) and tangent to the line x-5y=10.
1. radius of the circle
2. equation of a circle in a center-radius form
3. equation of circle in general form

Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
the circle passes through the points A(-5,2) and B(-1,4) and tangent to the line x-5y=10.
1. radius of the circle
2. equation of a circle in a center-radius form
3. equation of circle in general form
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Find the eqn of the perpendicular bisector of the line thru A & B. The center is on that line.
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The intersection of the perp bisector and x-5y=10 is a point on the circle, point C.
Find the perp bisector of AC.
The intersection of the 2 perp bisectors is the center.
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The radius is the distance from the center to A, B or C.
Find the

Answer by KMST(5396) About Me  (Show Source):
You can put this solution on YOUR website!
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
%284-2%29%2F%28-1-%28-5%29%29=2%2F4=1%2F2
The perpendicular bisector of AB is
y-3=-2%28x%2B3%29 --> 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%29%5E2%2B%28y-k%29%5E2=R%5E2 where R is the radius.
I can even substitute k=-2h-3 and get as equation for our circle
x-h%29%5E2%2B%28y%2B2h%2B3%29%5E2=R%5E2
Substituting the coordinates of B, I get an expression for R%5E2
-1-h%29%5E2%2B%284%2B2h%2B3%29%5E2=R%5E2 --> h%5E2%2B2h%2B1%2B%287%2B2h%29%5E2=R%5E2 --> h%5E2%2B2h%2B1%2B4h%5E2%2B28h%2B49=R%5E2 --> 5h%5E2%2B30h%2B50=R%5E2
Now I can write the equation for our circle as
x-h%29%5E2%2B%28y%2B2h%2B3%29%5E2=5h%5E2%2B30h%2B50
Simple. I just have to find h.
I know the circle is tangent to the line
x-5y=10 --> x=5y%2B10
so that line and the circle have just 1 intersection point.
Substituting x=5y%2B10 into the equation for the circle,
x-h%29%5E2%2B%28y%2B2h%2B3%29%5E2=5h%5E2%2B30h%2B50
I can find that intersection point and more.
%285y%2B10-h%29%5E2%2B%28y%2B2h%2B3%29%5E2=5h%5E2%2B30h%2B50
25y%5E2%2B100y-10hy%2Bh%5E2%2B100-20h%2By%5E2%2B6y%2B4h%2B4h%5E2%2B9%2B12h=5h%5E2%2B30h%2B50
26y%5E2%2B%28106-6h%29y%2B5h%5E2%2B109-8h=5h%5E2%2B30h%2B50
26y%5E2%2B%28106-6h%29y%2B59-38h=0
If that quadratic equation must have just one solution, the discriminant must be zero, so
%28106-6h%29%5E2-4%2A26%2A%2859-38h%29=0
11236-1272h%2B36h%5E2-6136%2B3952h=0
36h%5E2%2B2680h%2B5100=0
The best I can do with that unwieldy equation is divide everything by 4 to get
9h%5E2%2B670h%2B1275=0
Applying the quadratic formula
h+=+%28-670+%2B-+sqrt%28+670%5E2-4%2A9%2A1275+%29%29%2F%282%2A9%29+
h+=+%28-670+%2B-+sqrt%28+448900-45900+%29%29%2F18+
h+=+%28-670+%2B-+sqrt%28403000%29%29%2F18+
That can be simplified to
h+=+%28-335+%2B-+5sqrt%284030%29%29%2F9+
Substituting into k=-2h-3, we get
h+=+%28-335+%2B+5sqrt%284030%29%29%2F9+ with k=%28643-10sqrt%284030%29%29%2F9 and
h+=+%28-335+-+5sqrt%284030%29%29%2F9+ with k=%28643%2B10sqrt%284030%29%29%2F9
The approximate values give us points (-1.95,0.91) and (-72.49,141.98) for centers.
Exact values for R%5E2 are
R%5E2=%28978475-15400sqrt%284030%29%29%2F81 and R%5E2=%28978475%2B15400sqrt%284030%29%29%2F81
and approximate values are R=3.235 and R=155.4 corresponding to the centers above, respectively.
and
The equation of the circles are
for the small circle, and
for the large circle.
Asking for the equation of circle in general form is cruel.