Question

25) The endpoints of the diameter of a circle are (6, 5) and (−2, 3). Which equation represents the circle? A) (x − 2)2 + (y − 4)2 = 9 B) ( x+ 2)2 + (y + 4)2 = 16 C) (x − 2)2 − (y + 4)2 = 16 D) (x − 2)2 + (y − 4)2 = 17

Answer 1

Answer:

Option D.

Step-by-step explanation:

It is given that the endpoints of the diameter of a circle are (6, 5) and (−2, 3). So, midpoint of these end point is the center of the circle.

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

$Center=\left(\dfrac{6+(-2)}{2},\dfrac{5+3}{2}\right)$

$Center=\left(\dfrac{4}{2},\dfrac{8}{2}\right)$

$Center=\left(2,4\right)$

So, center of the circle is (2,4).

Length of diameter of the circle is the distance between (6, 5) and (−2, 3).

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

$d=\sqrt{(-2-6)^2+(3-5)^2}$

$d=\sqrt{(-8)^2+(-2)^2}$

$d=\sqrt{64+4}$

$d=\sqrt{68}$

$d=2\sqrt{17}$

So, radius of the circle is

$r=\dfrac{d}{2}=\dfrac{2\sqrt{17}}{2}=\sqrt{17}$

Standard form of a circle is

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

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

Substitute h=2, k=4 and $r=\sqrt{17}$ in the above equation.

$(x-2)^2+(y-4)^2=(\sqrt{17})^2$

$(x-2)^2+(y-4)^2=17$

Therefore, the correct option is D.