# Thread: find the difference between the centers of circles

1. ## find the difference between the centers of circles

Can someone tell me if this is right?
The equation of Circle P is x2 + y2 − 6x + 4y = 21 and the equation of circle Q is x2 + y2 − 8x + 10y = 11. What is the distance between the centers of these circles?

Circle p
(x-6)2+(y+4)2=21
x2+36+y2+16=21
x2+y2=21-36-16=-31
x2+y2=-31squared
Circle Q
x2+y2-8x+10y=11
(x-8)2+(y+10)2=11
x2+64+42+100=11
x2+y2=11-164=-153

2. Originally Posted by sanee66
Can someone tell me if this is right?
The equation of Circle P is x2 + y2 − 6x + 4y = 21 and the equation of circle Q is x2 + y2 − 8x + 10y = 11. What is the distance between the centers of these circles?

Circle p
(x-6)2+(y+4)2=21
x2+36+y2+16=21
x2+y2=21-36-16=-31
x2+y2=-31squared
Circle Q
x2+y2-8x+10y=11
(x-8)2+(y+10)2=11
x2+64+42+100=11
x2+y2=11-164=-153

No, if you had done everything else right that would be the difference in the radii of the two circles.

Circle P has equation: $\displaystyle x^2+y^2-6x+4y=21$.

Which after completing the square for $\displaystyle x$ and $\displaystyle y$ on the left hand side, can be rewritten as:

$\displaystyle (x-3)^2 + (y+2)^2=21-9-4= 8$

So the centre of circle P is at (3,-2).

Now do the same for circle Q, and then find the distance between the centres of the circles.

RonL

3. ## Did I get difference between the centers of circles

So is this right?
Circle P
x2+y2-6x+4y=21
x2-6x+y2+4y=21
x-3)2+(y+2)2=21
x2+9+y2+4=21
x2+y2-9-4=21-9-4=8
Center is at (3,-2
Circle Q
x2+y2-8x+10y=11
x2-8x+y2+10y=11
x-4)2+(y+5)2=11
x2+16+y2+25=11
x2+y2-16-25=11+16+25=52
Center is at (4,-5)
Distance between is (1,-3)

4. Originally Posted by CaptainBlack
$\displaystyle (x-3)^2 + (y+2)^2=21-9-4= 8$
Originally Posted by sanee66
x-3)2+(y+2)2=21
You didn't factor correctly

$\displaystyle (y+2)^2=(y+2)(y+2)=y^2+4y+4\neq y^2+4$

same thing with x-3