Pretty much, write two problems that equal to 0
The standard form for the equation of a circle is :
<span><span><span> (x−h)^</span>2</span>+<span><span>(y−k)^</span>2</span>=<span>r2</span></span><span> ----------- EQ(1)
</span><span> where </span><span>handk</span><span> are the </span><span>x and y</span><span> coordinates of the center of the circle and </span>r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :
((−10+(−8))/2,(1+5)/2)=(−9,3)
So the point (−9,3) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(−10−(−9))^2+(1−3)^2=1+4=5
⇒r=√5
Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :
(x+9)^2+(y−3)^2=5
Answer:
your answer is C.
Step-by-step explanation:
ANSWER:
STEP BY STEP EXPLANATION:
