# Thread: coordinate geometry regarding circle

1. ## coordinate geometry regarding circle

Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.

what i tried to do is:
$\displaystyle y=-2x+6 -----(1)$
let centre of circle be (x,y)
$\displaystyle (-2+x)^2+(0+y)^2=r^2$
$\displaystyle 4-4x+x^2+y^2=r^2 ------(2)$
$\displaystyle (4+x)^2+(6+y)^2=r^2$
$\displaystyle 16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$\displaystyle x=10$
Subst. x=10 into (1),
$\displaystyle y=-14$
centre of circle is (10,-14)..

but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??

2. Originally Posted by wintersoltice
Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.

what i tried to do is:
$\displaystyle y=-2x+6 -----(1)$
let centre of circle be (x,y)
$\displaystyle (-2+x)^2+(0+y)^2=r^2$
$\displaystyle 4-4x+x^2+y^2=r^2 ------(2)$
$\displaystyle (4+x)^2+(6+y)^2=r^2$
$\displaystyle 16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$\displaystyle x=10$
Subst. x=10 into (1),
$\displaystyle y=-14$
centre of circle is (10,-14)..

but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??
I would do it like this:

Let $\displaystyle (x_0,y_0)$ be a point on the line (in this case, also let it be the center of our circle).

Thus, the equation would be $\displaystyle (x-x_0)^2+(y-y_0)^2=r^2$

WLOG, let r be the distance from $\displaystyle (x_0,y_0)$ to $\displaystyle (-2,0)$: $\displaystyle r=\sqrt{(x_0+2)^2+y_0^2}$

So now it follows that $\displaystyle (x-x_0)^2+(y-y_0)^2=(x_0+2)^2+y_0^2$.

Now, if we plug in the other point, we have

$\displaystyle (4-x_0)^2+(6-y_0)^2=(x_0+2)^2+(y_0)^2\implies 16-8x_0+x_0^2+36-12y_0+y_0^2=x_0^2+4x_0+4+y_0^2$.

Simplifying, we have $\displaystyle 48=12x_0+12y_0\implies x_0+y_0=4$

Recall that $\displaystyle (x_0,y_0)$ is on the line $\displaystyle 2x+y=6\implies y_0=6-2x_0$.

Thus, $\displaystyle x_0+(6-2x_0)=4\implies -x_0+6=4\implies -x_0=-2\implies x_0=2$

Therefore, $\displaystyle y_0=6-2x_0=6-2(2)=6-4=2$.

Therefore, the center of the circle is $\displaystyle (2,2)$ and the equation now becomes $\displaystyle (x-2)^2+(y-2)^2=20$

Does this make sense?