# Thread: Find the equation and comman chord of a circle

1. ## Find the equation and comman chord of a circle

Find the equation and length of common chord of the circle
$
2x^2 + 2y^2 + 7x -5y + 2 = 0
$
and $
x^2 + y^2 - 4x + 8y - 18 =0
$

2. First, solve for y in each of the equations. Start by moving all x terms and constants to the right side

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

complete the square
divide by the coefficient of the quadratic term

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

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

Do the same with the other equation and solve the system.

3. Originally Posted by zorro
Find the equation and length of common chord of the circle
$
2x^2 + 2y^2 + 7x -5y + 2 = 0
$
and $
x^2 + y^2 - 4x + 8y - 18 =0
$
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 $\bigl(-\tfrac74,\tfrac54\bigr)$. The distance between these points is about 6.86. This is a lot more than the sum of the two radii, which are $\sqrt2$ and $\sqrt{58}/4$, both of them less than 2.)

Edit. Oops! First radius is $\sqrt{38}$, not $\sqrt2$. Thanks to skeeter for spotting that.

4. uhh ... the two circles do intersect.

5. Originally Posted by zorro
Find the equation and length of common chord of the circle
$
2x^2 + 2y^2 + 7x -5y + 2 = 0
$
and $
x^2 + y^2 - 4x + 8y - 18 =0
$

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

$-2(x^2 + y^2 - 4x + 8y - 18 =0)$

---------------------------------

$15x - 21y + 38 = 0$

linear equation is the equation of the chord

6. what about the length of the chord ????

7. Originally Posted by zorro
what about the length of the chord ????
You know the equation of the chord and so can find the points where that chord intersects either of the circles (it intersects both circles in the same points, of course). Find the distance between those two points.