Question

The endpoints of a diameter of a circle are located at(-3,6) and (7,-12). What is the equation of the circle? A. (×+2)2+(y+3)2=square 106 B. (X-4)2+(y+6)2

Answer 1

Answer:

$(x-2)^2+(y+3)^2=106$.

Step-by-step explanation:

It is given that the endpoints of a diameter of a circle are located at(-3,6) and (7,-12). It means center of the circle is the midpoint of (-3,6) and (7,-12).

$Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Center=\left(\dfrac{-3+7}{2},\dfrac{6-12}{2}\right)$

$Center=\left(2,-3\right)$

The center of the circle is (2,-3).

Radius is the distance between center and the endpoint of diameter. So, radius is equal to the distance between (2,-3) and (-3,6).

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

$r=\sqrt{25+81}$

$r=\sqrt{106}$

The standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center and r is radius.

Substitute h=2, k=-3 and $r=\sqrt{106}$ in the above equation.

$(x-2)^2+(y-(-3))^2=(\sqrt{106})^2$

$(x-2)^2+(y+3)^2=106$

Therefore, the required equation of circle is $(x-2)^2+(y+3)^2=106$.