# Thread: Coordinates Geometry of circle

1. ## Coordinates Geometry of circle

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?

2. Originally Posted by helloying
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]: $x^2+y^2-2x-3=0$

[2]: $x^2+y^2+x-y-6=0$

Calculate [2] - [1]:

$3x+2y-6=0~\implies~y=-\dfrac32x+3$ Substitute the term for y into equation [1]:

$x^2+\left(-\dfrac32x+3 \right)^2-2x-3\left(-\dfrac32x+3\right)=0~\implies ~ \dfrac{13}4x^2-\dfrac{13}2x=0\implies~x=0~\vee~x=2$

That means the coordinates of P and Q are: $P(0,3), Q(2,0)$

Now use 2-point-formula of a straight line to get the equation of PQ.

Use distance formula to calculate the length of $\overline{PQ}$

EDIT: See attachment