Question

the equation of circle through the intersection of circles x^2+y^2-3x-6y+8=0 and x^2+y^2-2x-4y+4=0 and touching the line x+2y=5?

Answer 1

@agent0smith

Answer 2

for convenience, I give out what I got:
the intersection points are on the line \(y = -\dfrac{1}{2}x +2\) and it // with the line x+2y =5

Answer 3

since it touches the line x +2y =5, the equation of the line which link the center of the required circle and the touching point has the slope m =2
then I am stuck,

Answer 4

see the graph

Answer 5

those two circles have two intersections, and the point at (0,2) also coincides with the line y=-1/2x+2.
So the origin of the new circle should be at (0,2)

Answer 6

Seems to be a bit of a brute force way to me (the one I am about to suggest), furthermore I haven't tried it yet because it involves quite a few calculations. But if you manage to find the two points of intersections this will give you two conditions for your circle equation, the third one would be that the distance between the line to the midpoint of your circle is \(r\). Use Hesse Normal Form for that.

\[\Large \left|\frac{x_0+2y_0-5}{\sqrt{5}}\right|=r \]

Answer 7

Using this brute force way of solving it I obtained the following result: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2-3x-6y%2B8%3D0%2C+x%5E2%2By%5E2-2x-4y%2B4%3D0%2C+%28x-%281%2F2%29%29%5E2%2B%28y-1%29%5E2%3D%286.25%2F5%29%2C+x%2B2y%3D5

Answer 8

You can solve it by plotting

Answer 9

so I came out with \(x_0 = \frac{1}{2}\) and \(y_0=1\) with radius squared \(r^2= \frac{5}{4}\), not the most elegant approach though.

Answer 10

To summarize my approach:

1. Find points of intersections \(P_1\) and \(P_2\) giving you two conditions for your unknown circle with general equation \((x-x_0)^2+(y-y_0)^2=r^2\)
2. Set the radius equal to the distance from the line to the unknown center \(x_0,y_0\) using Hesse Normal Form of a line equation.

Solve, solve solve. Or even better, let a computer do it.

Answer 11

we can find 3 points on the new circle: 2 from the intersection of the given circles, the third from the intersection of the perpendicular chord bisector and the given tangent line.
once you have 3 points you can find the circle.

Answer 12

In general, it will be a lot of algebraic manipulation (-:

Answer 13

Thank you everybody.
@Spacelimbus If there is only one way to do and we cannot avoid calculating, I give up. hihihi, since it 's not my problem.
@phi I thought of that method but I foresaw a bunch of things to do on that way. hihihi.... That was the reason I tag others to find the shortcut. :)