Respuesta :
Answer:
The equation of [tex]C_{1}[/tex] is: [tex]x^2+y^2=4[/tex]
and the equation of [tex]C_{2}[/tex] is: [tex]x^2+y^2-2\sqrt{2}x-2\sqrt{2}y+2=0[/tex]
Step-by-step explanation:
The standard form of circle equation: [tex](x-h)^2+(y-k)^2 = r^2[/tex] , where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius of the circle.
The center of the circle [tex]C_{1}[/tex] is at origin or at (0, 0) and has radius of 2.
So, the equation of the circle [tex]C_{1}[/tex] will be........
[tex](x-0)^2+(y-0)^2=(2)^2\\ \\ x^2+y^2=4 ..........................................(1)[/tex]
Now, the center of [tex]C_{2}[/tex] is the point in the first quadrant, where the line [tex]y=x[/tex] intersects [tex]C_{1}[/tex]
Solving the equations [tex]x^2+y^2=4[/tex] and [tex]y=x[/tex] , we will get......
[tex]x^2+x^2= 4[/tex] (Substituting [tex]y[/tex] as [tex]x[/tex])
[tex]2x^2= 4\\ \\ x^2=2\\ \\ x=\pm \sqrt{2}[/tex]
So, [tex]y=x = \pm \sqrt{2}[/tex]
Thus, the center of the circle [tex]C_{2}[/tex] will be at [tex](\sqrt{2},\sqrt{2})[/tex] (Negative value is ignored as the point is in first quadrant)
Now, the distance between the center of [tex]C_{2}[/tex] and the x-axis is [tex]\sqrt{2}[/tex] and the circle touches the x-axis only at one point.
That means, the radius of the circle [tex]C_{2}[/tex] will be also [tex]\sqrt{2}[/tex]
So, the equation of the circle [tex]C_{2}[/tex] will be......
[tex](x-\sqrt{2})^2+(y-\sqrt{2})^2= (\sqrt{2})^2\\ \\ x^2-2\sqrt{2}x+2+y^2-2\sqrt{2}y+2=2\\ \\ x^2+y^2-2\sqrt{2}x-2\sqrt{2}y+2=0[/tex]
