two circles, x²+y²-2x-3y=0,
x²+y²+x-y-6=0
intersect at points P and Q. Find the length of PQ and eq of the line PQ.
Tell me wat i shld do?
[1]: $\displaystyle x^2+y^2-2x-3=0$
[2]: $\displaystyle x^2+y^2+x-y-6=0$
Calculate [2] - [1]:
$\displaystyle 3x+2y-6=0~\implies~y=-\dfrac32x+3$ Substitute the term for y into equation [1]:
$\displaystyle x^2+\left(-\dfrac32x+3 \right)^2-2x-3\left(-\dfrac32x+3\right)=0~\implies ~ \dfrac{13}4x^2-\dfrac{13}2x=0\implies~x=0~\vee~x=2$
That means the coordinates of P and Q are: $\displaystyle P(0,3), Q(2,0)$
Now use 2-point-formula of a straight line to get the equation of PQ.
Use distance formula to calculate the length of $\displaystyle \overline{PQ}$
EDIT: See attachment