1. ## Geometry: Circle

$x^2+y^2-8x-4y-5=0$
and
$x^2+y^2+10x+20y+25=0$
Deduce that the two circles touch each other. If the line of centres meets the circles in points P,Q,R, find the equation of the circle that has PR as diameter.

r1+r2=15
sq root of (9^2+12^2) =15

therefore we can say they touch. but dont understand the second bit.
I don't know how to find the equation from the diameter of 15 from PR

PLease show step by step working!

Many thanks!

2. Originally Posted by BabyMilo
therefore we can say they touch. but dont understand the second bit.
$radius\ of\ circle\ having\ PQ\ diameter\ = r_{1}=5$
$radius\ of\ circle\ having\ QR\ diameter\ = r_{2}=10$
$radius\ of\ circle\ having\ PR\ diameter\ = r_{1}+r_{2}=15$
centre of circle having PR diameter = mid points S(x,y) of PR

3. but the answer is $(x+2)^2+(y+6)^2=15^2$

4. Originally Posted by BabyMilo
PLease show step by step working!
NO! You do the work. Show it to us. We can help you then.
First find the center of each circle.
Find the distance between them.
Is the distance the sum of the radii?

5. Originally Posted by Plato
NO! You do the work. Show it to us. We can help you then.
First find the center of each circle.
Find the distance between them.
Is the distance the sum of the radii?
Calm down! I've already worked out that part as i stated in the first thread
r1+r2=15
sq root of (9^2+12^2) =15
C1=(4,2) r=5
C2=(-5,-10) r=10
The big Circle radii = 15
diameter=30

I wanted you guys to show step by step working because i could understand what you were doing that's all, not me being lazy.