Question

The equation of circle having a diameter with endpoints (-2, 1) and (6, 7) is

Answer 1

Answer:

$(x-2)^{2}+(y-4)^{2}=25$

Step-by-step explanation:

step 1

Find the diameter of the circle

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

$A(-2,1)\\B(6,7)$  

substitute the values

$d=\sqrt{(7-1)^{2}+(6+2)^{2}}$

$d=\sqrt{(6)^{2}+(8)^{2}}$

$d=\sqrt{100}$

$d=10\ units$

The radius is half the diameter

so

$r=10/2=5\ units$

step 2

Find the center of the circle

the center of the circle is the midpoint between the endpoints of the diameter

so

The center is

$(\frac{-2+6}{2},\frac{1+7}{2})$

$(2,4)$

step 3

Find the equation of the circle

The equation of the circle is

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

substitute the values

$(x-2)^{2}+(y-4)^{2}=5^{2}$

$(x-2)^{2}+(y-4)^{2}=25$