Find the condition on a b c such that two chords of the circle x^2+y^2-2ax-2by+a^2+b^2-c^2=0 passing through the point (a,b+c) are bisected by the line x=y
Answers
Answer:
a ≠ b + c
and
c² > 4 ( a - b ) ( a - b - c )
Step-by-step explanation:
- The point (a, b+c) is on the circumference of the circle (just put x = a and y = b+c in the equation and confirm that it is identically zero).
- Given a point A on a circle C and a line L, let A' be the reflection of A about L and let M be the line through A' parallel to L. Let S and T be the points where M meets C. Then the chords AS and AT are bisected by L and these are the only such chords.
- So the requirements are that the line M meets the C in two points (so that there are two chords) and A should not be on L (that would make M' pass through A so A = S or A = T).
- In our case, we have A = (a, b+c) and L is the line y=x. The requirement that A not be on L is then simply a ≠ b+c.
- For our A and L, the point A' = (b+c, a). The line M has the same slope as y=x and passes through A', to it is given by y = x + a - b - c.
- As the points S and T are on the circle C as well as the line M, they satisfy the equations for both C and M. Substituting y = x + a - b - c into the equation for the circle and doing some tidying up, the resulting equation for x is:
x² - ( 2b + c ) x + a² + 2b² - 2ab + 2bc - ac = 0.
- This equation has two solutions if and only if the discriminant is positive. The discriminant is
( 2b + c )² - 4 ( a² + 2b² - 2ab + 2bc - ac )
= c² - 4 ( a - b ) ( a - b - c ).
Requiring this to be positive is equivalent to
c² > 4 ( a - b ) ( a - b - c ).