Answer :
The equation of the circle is [tex](x+3)^{2}+(y-3)^{2}=37[/tex]
Explanation:
Given that the endpoints of the circle are A(-4,9) and B(-2,-3)
We need to determine the equation of the circle.
Center:
The center of the circle can be determined using the midpoint formula,
[tex]Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]
Substituting the coordinates A(-4,9) and B(-2,-3), we get,
[tex]Center=(\frac{-4-2}{2},\frac{9-3}{2})[/tex]
[tex]Center=(\frac{-6}{2},\frac{6}{2})[/tex]
[tex]Center=(-3,3)[/tex]
Thus, the center of the circle is (-3,3)
Radius:
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 (-3,3) and the endpoint (-4,9), we get,
[tex]r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}[/tex]
[tex]r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}[/tex]
[tex]r=\sqrt{1+36}[/tex]
[tex]r=\sqrt{37}[/tex]
Thus, the radius of the circle is [tex]\sqrt{37}[/tex]
Equation of the circle:
The standard form of the equation of the circle is given by
[tex](x-a)^{2}+(y-b)^{2}=r^{2}[/tex]
where (a,b) is the center and r is the radius.
Substituting the values, we have,
[tex](x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}[/tex]
[tex](x+3)^{2}+(y-3)^{2}=37[/tex]
Thus, the equation of the circle is [tex](x+3)^{2}+(y-3)^{2}=37[/tex]