Question

Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )

Answer 1

The equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$

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,

$Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

$Center=(\frac{-4-2}{2},\frac{9-3}{2})$

$Center=(\frac{-6}{2},\frac{6}{2})$

$Center=(-3,3)$

Thus, the center of the circle is (-3,3)

Radius:

The radius of the circle can be determined using the distance formula,

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substituting the center (-3,3) and the endpoint (-4,9), we get,

$r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}$

$r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}$

$r=\sqrt{1+36}$

$r=\sqrt{37}$

Thus, the radius of the circle is $\sqrt{37}$

Equation of the circle:

The standard form of the equation of the circle is given by

$(x-a)^{2}+(y-b)^{2}=r^{2}$

where (a,b) is the center and r is the radius.

Substituting the values, we have,

$(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}$

$(x+3)^{2}+(y-3)^{2}=37$

Thus, the equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$