Find the equation and length of common chord of the circle

$\displaystyle

2x^2 + 2y^2 + 7x -5y + 2 = 0

$ and $\displaystyle

x^2 + y^2 - 4x + 8y - 18 =0

$

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- May 3rd 2009, 10:31 PMzorroFind the equation and comman chord of a circle
Find the equation and length of common chord of the circle

$\displaystyle

2x^2 + 2y^2 + 7x -5y + 2 = 0

$ and $\displaystyle

x^2 + y^2 - 4x + 8y - 18 =0

$ - May 5th 2009, 10:37 AMVonNemo19
First, solve for y in each of the equations. Start by moving all x terms and constants to the right side

$\displaystyle 2y^2-5y=-2x^2-7x-2$

complete the square

divide by the coefficient of the quadratic term

$\displaystyle (y-\frac{5}{4})^2=-x^2-\frac{7}{2}x-1+\frac{25}{16}$

taking the root and adding

$\displaystyle y=\pm{\sqrt{-2x^2-\frac{7}{2}x-1+\frac{25}{16}}}+\frac{5}{4}$

Do the same with the other equation and solve the system. - May 5th 2009, 12:45 PMOpalg
Something wrong here. Those two circles don't intersect at all, so how can they have a common chord? (The centres of the circles are at (2,–4) and $\displaystyle \bigl(-\tfrac74,\tfrac54\bigr)$. The distance between these points is about 6.86. This is a lot more than the sum of the two radii, which are $\displaystyle \sqrt2$ and $\displaystyle \sqrt{58}/4$, both of them less than 2.)

**Edit.**Oops! First radius is $\displaystyle \sqrt{38}$, not $\displaystyle \sqrt2$. Thanks to skeeter for spotting that. - May 5th 2009, 03:18 PMskeeter
uhh ... the two circles do intersect.

- May 5th 2009, 03:26 PMskeeter
- Dec 27th 2009, 11:51 PMzorro
what about the length of the chord ????

- Dec 28th 2009, 03:24 AMHallsofIvy