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Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.

So far I have:$$(x-h)^2 + (y-k)^2=r^2$$ Since the center is on the line $x=4$, then the center is $(4;k)$.
$\therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=r^2$
And because the circle touches the $x-axis$, $r=k$.
$\therefore~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=k^2$

Now, I am struggling to use $x^2 + y^2=4$ to find $k$.

Please help.

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This image is enough to understand the answer.

$$x^2+y^2=0$$

this circle have radius $=2$
center$=(0,0)$
So, in $\triangle OO'A$ $$(OO')^2=(O'A)^2+(OA)^2\tag{Pythagoras theorem}$$ $$(OB+BO')^2=(O'A)^2+(OA)^2$$ $\therefore$ $$(2+k)^2=k^2+4^2$$ $$4+k^2+4k=k^2+16$$ $\therefore$ $$k=3$$

Hope it helps!!!

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let $$B(xB,yB)$$ the common Point of the two circles then we have $$xB^2+yB^2=4$$ $$(xB-4)^2+(yB-yM)^2=yM^2$$ solving this System we get $$\frac{10-4xB)^2}{yM}=yB$$ from here we get the equation $$xB^2+\left(\frac{10-4xB}{y_M}\right)^2=4$$ the dicriminant of this equation must be Zero, thus we have $$yM^2(yM^2-9)=0$$ and we get $$yM=\pm 3$$

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