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Sagot :
Given:
The center of the circle is the point ( 3, 2 )
And the point on the circle is ( 4, 3 )
To write the equation of the circle, we need to find the radius
The radius = the distance between the center and the point on the circle
so, the radius is the distance between ( 3, 2) and ( 4, 3)
So,
[tex]\begin{gathered} r=d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ r=\sqrt[]{(4-3)^2+(3-2)^2^{}}=\sqrt[]{1+1}=\sqrt[]{2} \end{gathered}[/tex]The general equation of the circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]Where (h, k) is the coordinates of the center of the circle, r is the radius
So,
[tex]\begin{gathered} (h,k)=(3,2) \\ r=\sqrt[]{2} \end{gathered}[/tex]so, the equation of the circle will be:
[tex](x-3)^2+(y-2)^2=2[/tex]
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