Question

what is the equation of a circle wit diameter AB that has endpoint A(0,0) and B(8,6)
a.(x-3)^2+(y-4)=5
b. (x-4)^2+(Y-3)^2=25
c.(x-4)^2+(y-3)^2=5
d.(x-3)^2+(y-4)^2=25

Answer 1

That diameter is a line whose equation is y=3/4 x + 0
The midpoint is P(4,3) of the line and also the center of the circle.
The radius would be:\[\sqrt{3^{2}+4^{2}}=5\]
Do you have enough data now to complete the equation for the circle?

Answer 2

For a circle whose center is located at P(4,3) then h=4, and k=3 and radius 5 would have an equation:\[(x-4)^{2}+(y-3)^{2}=25\] the 25 is the radius squared.
working that equation would give you:\[x ^{2}-8x+16+y ^{2}-6y+9=25\]

Expanding and simplifying we get
\[x^{2}+y ^{2}-8x-6y=0\]

Answer 3

I think!