coordinate geometry regarding circle

**wintersoltice** · September 22nd 2009, 03:35 AM

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:
$y=-2x+6 -----(1)$
let centre of circle be (x,y)
$(-2+x)^2+(0+y)^2=r^2$
$4-4x+x^2+y^2=r^2 ------(2)$
$(4+x)^2+(6+y)^2=r^2$
$16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$x=10$
Subst. x=10 into (1),
$y=-14$
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??

**Chris L T521** · September 22nd 2009, 04:27 AM

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:
$y=-2x+6 -----(1)$
let centre of circle be (x,y)
$(-2+x)^2+(0+y)^2=r^2$
$4-4x+x^2+y^2=r^2 ------(2)$
$(4+x)^2+(6+y)^2=r^2$
$16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$x=10$
Subst. x=10 into (1),
$y=-14$
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 $(x_0,y_0)$ be a point on the line (in this case, also let it be the center of our circle).

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

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

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

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

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

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

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

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

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

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

Does this make sense?

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