Sagot :
✏️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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