Question

The points J(-1,-9) and K(5,1) are endpoints of a diameter of circle S. Which equation represents circle S? A. (x − 2)2 + (y + 4)2 = 34 B. (x − 2)2 + (y + 4)2 = 136 C. (x + 2)2 + (y − 4)2 = 34 D. (x + 2)2 + (y − 4)2 = 136

Answer 1

Answer:

the equation of the circle is

$(x-2)^2+(y+4)^2 = 34$

Step-by-step explanation:

Recall that the equation of a circle is given by $(x-h)^2+(y-k)^2 = r^2$ where r is the radius, and (h,k) is the center. Recall that given two points that are the endpoints of a diameter, the center of the circle is their correspondent midpoint. Also, recall that given points (a,b), (c,d) their midpoint is obtained by $(\frac{a+c}{2},\frac{b+d}{2})$

In this case we are given the endpoints (-1,-9), (5,1). So the center of the circle is the midpoint obtained by $(\frac{-1+5}{2},\frac{-9+1}{2}) = (2, -4)$. Recall that the radius of a circle is the distance from the radius to any of the points of the circle. So, the radius is the distance between the center and the point (-1,-9). We will calculate r^2, by using the distance formula. REcall that the distance between points (a,b), (c,d) is given by

$\sqrt[]{(a-c)^2+(b-d)^2}$. So, its square is

$(a-c)^2+(b-d)^2$.

In our case,

$r^2 = (2-(-1))^2+(-4-(-9))^2 = 3^2+5^2 = 34$

So, the equation of the circle is

$(x-2)^2+(y+4)^2 = 34$