(x−h)^2+(y−k)^2=r2 ----------- EQ(1)
 where handk are the x and y coordinates of the center of the circle and r is the radius.
 The center of the circle is the midpoint of the diameter.

So the midpoint of the diameter with endpoints at (2,-5)and(8,-9) is :

((2+(8))/2,(-5+(-9))/2)=(5,-7)

So the point (5,-7) is the center of the circle.

Now, use the distance formula to find the radius of the circle:

r^2=(2−(5))^2+(-5−(-7))^2=9+4=13

⇒r=√13

Subtituting h=5, k=-7 and r=√13 into EQ(1) gives :

(x-5)^2+(y+7)^2=13

5 0

4 years ago

Evaluate b-(-1/8)+c where b=2 and c=-7/-4

Kay [80]

Answer:

Just plug in the values and the calculate :

$\longrightarrow\tt2 - ( - \frac{1}{8}) + ( \frac{ - 7}{ - 4} )$

$\longrightarrow \tt2 + \frac{1}{8} + \frac{7}{4}$

$\longrightarrow \boxed{\tt \frac{31}{8} }$

----------- HappY LearninG <3 -----------

3 0

2 years ago