What is the equation for the circle, if it has diameter endpoints at (3,-4) and (-5,-4)?

(x+1)^2+(y+4)^2=16

Step-by-step explanation:

equation of the circle

(x-x0)^2+(y-y0)^2=r^2

where x0 and y0 is center of circle

and r is radius of the circle.

we know that the center point is the middle of the diameter

so x0=(x1+x2)/2

x0=(3+(-5))/2=-2/2=-1

y0=(-4+(-4))/2=-8/2=-4

so the midpoint or the center of circle is the points (-1,-4)

the formula to find the distance between two. points is √[(x2-x1)^2+(y2-y1)^2]

so r is the distance between the center point and one of the diameter endpoints.

let us take points (3,-4) and center point (-1,-4)

√(-1-3)^2+(-4-(-4))^2=

√(-4)^2+0^2=

√16=4

so the radius of the circle is 4

now the equation of the circle will be

(x-(-1))^2+(y-(-4))^2=4^2

(x+1)^2+(y+4)^2=16

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23