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

- Thread starter Ryousekki
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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.

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

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Both circles share a common centre \(\displaystyle (0,0)\)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)

Can they intersect? EXPLAIN!

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\(\displaystyle y^{2}=36-x^{2}\iff y=\pm\sqrt{36-x^{2}}\iff f(x)=\pm\sqrt{36-x^{2}}\)

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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?

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\(\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.

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