raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the circle. Write the equation of the circle.
Accepted Solution
A:
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-versaWhen 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.