Question

If (5, 4) and (7, 2) are the endpoints of a diameter of a circle, what is the equation of the circle?

Answer 1

The equation of the circle is $(x-6)^{2}+(y-3)^{2}=2$

Explanation:

The endpoints of the diameter of a circle are $(5,4)$ and $(7,2)$

We need to determine the equation of the circle.

The center of the circle can be calculated using midpoint formula.

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

Thus, we have,

Center = $(\frac{5+7}{2} ,\frac{4+2}{2} )$

            $=(\frac{12}{2} ,\frac{6}{2} )$

            $=(6,3)$

Thus, the center of the circle is $(6,3)$

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 $(6,3)$ and the endpoint $(5,4)$, we have,

$r=\sqrt{\left(5-6\right)^{2}+\left(4-3\right)^{2}}$

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

$r=\sqrt{1+1}$

$r=\sqrt{2}$

Thus, the radius of the circle is $\sqrt{2}$

The standard form of the equation of the circle is

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

Substituting $(6,3)$ and $r=\sqrt{2}$

Thus, the equation of the circle becomes

$(x-6)^{2}+(y-3)^{2}=(\sqrt{2} )^2$

$(x-6)^{2}+(y-3)^{2}=2$

Therefore, the equation of the circle is $(x-6)^{2}+(y-3)^{2}=2$