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what is the equation of the circle with diameter whose endpoints are (3,1) and (5.5)

Sagot :

UK2019

✏️CIRCLE EQUATION

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[tex]\underline{\mathbb{QUESTION:}}[/tex]

  • What is the equation of the circle with diameter whose endpoints are (3,1) and (5,5)?

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[tex]\underline{\mathbb{ANSWER:}}[/tex]

[tex]\quad\Large\rm»\:\: \green{(x-4)^2+(y-3)^2=5}[/tex]

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[tex]\underline{\mathbb{SOLUTION:}}[/tex]

- The equation of the circle in standard form is written as:

  • [tex](x-h)^2+(y-k)²=r^2[/tex]

- Where (h,k) is the center and r is the radius.

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- Find the midpoint between the endpoints because that would be the center of the circle.

[tex] \begin{aligned}& \bold{ \color{lightblue}Formula:} \\& \boxed{M = \bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)}\end{aligned}[/tex]

  • [tex] \begin{aligned}{Center = \bigg(\frac{3 + 5}{2},\frac{1 + 5}{2}\bigg)}\end{aligned}[/tex]

  • [tex] \begin{aligned}{Center = \bigg(\frac{8}{2},\frac{6}{2}\bigg)}\end{aligned}[/tex]

  • [tex]Center = (4, 3)[/tex]

- The center is at (4,3). Substitute in the standard form of the equation.

  • [tex](x - 4)^{2} + (y - 3)^{2} = {r}^{2} [/tex]

- Find the square of the radius if it passes through one of the given endpoints of the diameter: (5,5)

  • [tex](5 - 4)^{2} + (5 - 3)^{2} = {r}^{2} [/tex]

  • [tex](1)^{2} + (2)^{2} = {r}^{2} [/tex]

  • [tex]1 + 4 = {r}^{2} [/tex]

  • [tex]5 = {r}^{2} [/tex]

- Thus, the radius² is 5. Substitute the square of the radius to the equation.

  • [tex](x - 4)^{2} + (y - 3)^{2} = 5[/tex]

[tex]\therefore[/tex] (x - 4)² + (y - 3)² = 5 is the standard form of the equation.

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