Respuesta :
The equation of the circle is [tex](x-6)^{2}+(y-3)^{2}=2[/tex]
Explanation:
The endpoints of the diameter of a circle are [tex](5,4)[/tex] and [tex](7,2)[/tex]
We need to determine the equation of the circle.
The center of the circle can be calculated using midpoint formula.
Midpoint = [tex](\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )[/tex]
Thus, we have,
Center = [tex](\frac{5+7}{2} ,\frac{4+2}{2} )[/tex]
[tex]=(\frac{12}{2} ,\frac{6}{2} )[/tex]
[tex]=(6,3)[/tex]
Thus, the center of the circle is [tex](6,3)[/tex]
The radius of the circle can be determined using the distance formula,
[tex]r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]
Substituting the center [tex](6,3)[/tex] and the endpoint [tex](5,4)[/tex], we have,
[tex]r=\sqrt{\left(5-6\right)^{2}+\left(4-3\right)^{2}}[/tex]
[tex]r=\sqrt{\left(-1\right)^{2}+\left(1\right)^{2}}[/tex]
[tex]r=\sqrt{1+1}[/tex]
[tex]r=\sqrt{2}[/tex]
Thus, the radius of the circle is [tex]\sqrt{2}[/tex]
The standard form of the equation of the circle is
[tex](x-a)^{2}+(y-b)^{2}=r^{2}[/tex]
Substituting [tex](6,3)[/tex] and [tex]r=\sqrt{2}[/tex]
Thus, the equation of the circle becomes
[tex](x-6)^{2}+(y-3)^{2}=(\sqrt{2} )^2[/tex]
[tex](x-6)^{2}+(y-3)^{2}=2[/tex]
Therefore, the equation of the circle is [tex](x-6)^{2}+(y-3)^{2}=2[/tex]
