Question

If (5, 4) and (7, 2) are the endpoints of a diameter of a circle, what is the equation of the circle?

Answer 1

The equation of the circle is
`(x-6)^(2)+(y-3)^(2)=2`

Step-by-step explanation:

The endpoints of the diameter of a circle are
`(5,4)` and
`(7,2)`

We need to determine the equation of the circle.

The center of the circle can be calculated using midpoint formula.

Midpoint =
`((x_1+x_2)/(2) ,(y_1+y_2)/(2) )`

Thus, we have,

Center =
`((5+7)/(2) ,(4+2)/(2) )`

`=((12)/(2) ,(6)/(2) )`

`=(6,3)`

Thus, the center of the circle is
`(6,3)`

The radius of the circle can be determined using the distance formula,

`r=\sqrt{\left(x_(2)-x_(1)\right)^(2)+\left(y_(2)-y_(1)\right)^(2)}`

Substituting the center
`(6,3)` and the endpoint
`(5,4)`, we have,

`r=\sqrt{\left(5-6\right)^(2)+\left(4-3\right)^(2)}`

`r=\sqrt{\left(-1\right)^(2)+\left(1\right)^(2)}`

`r=√(1+1)`

`r=√(2)`

Thus, the radius of the circle is
`√(2)`

The standard form of the equation of the circle is

`(x-a)^(2)+(y-b)^(2)=r^(2)`

Substituting
`(6,3)` and
`r=√(2)`

Thus, the equation of the circle becomes

`(x-6)^(2)+(y-3)^(2)=(√(2) )^2`

`(x-6)^(2)+(y-3)^(2)=2`

Therefore, the equation of the circle is
`(x-6)^(2)+(y-3)^(2)=2`