Answer:
Equation:
[tex]{x}^{2} + {y}^{2} + 2x - 2y - 35= 0[/tex]
The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.
Step-by-step explanation:
We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).
The center of this circle is the midpoint of (-3, 4) and (5, -2).
We use the midpoint formula:
[tex]( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )[/tex]
Plug in the points to get:
[tex]( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )[/tex]
[tex]( \frac{ -2}{2}, \frac{ 2}{2} )[/tex]
[tex]( - 1, 1)[/tex]
We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:
[tex]r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} } [/tex]
[tex]r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} } [/tex]
[tex]r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} } [/tex]
[tex]r = \sqrt{ 36+ 1 } = \sqrt{37} [/tex]
The equation of the circle is given by:
[tex](x-h)^2 + (y-k)^2 = {r}^{2} [/tex]
Where (h,k)=(-1,1) and r=√37 is the radius
We plug in the values to get:
[tex](x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2} [/tex]
[tex](x + 1)^2 + (y - 1)^2 = 37[/tex]
We expand to get:
[tex] {x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37[/tex]
[tex]{x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0[/tex]
[tex]{x}^{2} + {y}^{2} + 2x - 2y - 35= 0[/tex]
We want to find at least four points on this circle.
We can choose any point for x and solve for y or vice-versa
When y=0,
[tex]{x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0[/tex]
[tex]{x}^{2} +2x - 35= 0[/tex]
[tex](x - 5)(x + 7) = 0[/tex]
[tex]x = 5 \: or \: x = - 7[/tex]
The point (5,0) and (-7,0) lies on the circle.
When x=0
[tex]{0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0[/tex]
[tex] {y}^{2} - 2y - 35= 0[/tex]
[tex](y - 7)(y + 5) = 0[/tex]
[tex]y = 7 \: or \: y = - 5[/tex]
The point (0,-5) and (0,7) lie on this circle.
