Intersection of two circles

Ryousekki

New member
Is it possible to find the intersection of two circles eith these two equation
x^2+y^2=36
x^2+y^2=25

Based on my understanding this circles is both on the center which is point (0,0)

If its possible can you show also how to solve this im quite stuck with this one

lev888

Full Member
Is it possible to find the intersection of two circles eith these two equation
x^2+y^2=36
x^2+y^2=25

Based on my understanding this circles is both on the center which is point (0,0)

If its possible can you show also how to solve this im quite stuck with this one
Circles with the same center (concentric) are either identical or don't intersect. These 2 have different diameters, so no intersection.

pka

Elite Member
Is it possible to find the intersection of two circles eith these two equation
$$\displaystyle C_1:~x^2+y^2=36$$
$$\displaystyle C_2:~x^2+y^2=25$$
Based on my understanding this circles is both on the center which is point (0,0)
Both circles share a common centre $$\displaystyle (0,0)$$ BUT the radius of $$\displaystyle C_1$$ is $$\displaystyle 6$$ and the radius of $$\displaystyle C_2$$ is $$\displaystyle 5$$.
Can they intersect? EXPLAIN!

OmarMohamedKhallaf

New member
As lev888 said, logically the two circles don't intersect. If you want to try your self with equations, then the first circle is:
$$\displaystyle y^{2}=36-x^{2}\iff y=\pm\sqrt{36-x^{2}}\iff f(x)=\pm\sqrt{36-x^{2}}$$

pka

Elite Member
As lev888 said, logically the two circles don't intersect. If you want to try your self with equations, then the first circle is: $$\displaystyle y^{2}=36-x^{2}\iff y=\pm\sqrt{36-x^{2}}\iff f(x)=\pm\sqrt{36-x^{2}}$$
To: OmarMohamedKhallaf please tell us why if you know that is an expression is $$\displaystyle (x-h)^2+(y-k)^2=r^2$$ is a circle then that circle ha centre $$\displaystyle (h,k)$$ and radius $$\displaystyle |r|$$ if two circles have the same centre but different radii cannot intersect?

OmarMohamedKhallaf

New member
I don't understand exactly what is the meant from asking me this question, but no problem here's how I tried to solve it:
$$\displaystyle \\ \because r_{1}\text{ is the radius of }C_{1},\,r_{2}\text{ is the radius of }C_{2},\,r_{1}\ne r_{2}\\ \text{Suppose the two circles } (x-h)^{2}+(y-k)^{2}=r_{1}^{2},\,(x-h)^{2}+(y-k)^{2}=r_{2}^{2}\text{ intersect} \\ \therefore (x-h)^{2}+(y-k)^{2}=(x-h)^{2}+(y-k)^{2}\\ \text{but } r_{1}^{2}\ne r_{2}^{2} \\ \therefore (x-h)^{2}+(y-k)^{2}\ne (x-h)^{2}+(y-k)^{2} \\ \therefore \text{The two circles doesn't intersect}$$
If there's any thing wrong, kindly correct me.

Ryousekki

New member
Thank you very much for answering my question

HallsofIvy

Elite Member
Note the distinction between "circles" and "disks". The circles, $$\displaystyle x^2+ y^2= 36$$ and $$\displaystyle x^2+ y^2= 25$$, do not intersect. In fact, the shortest distance between any point in one circle and any point in the other is 1. The two disks. $$\displaystyle x^2+ y^2\le 36$$ and $$\displaystyle x^2+ y^2\le 25$$ have the smaller disk as intersection.