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Sagot :
✒️CIRCLE EQUATION
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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad \large \:\: \rm (x-4)^2 + (y-1)^2 = 50 [/tex]
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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]
The center of the circle is the midpoint of the diameter segment. Find the midpoint between the endpoints of the diameter.
[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm Midpoint = \bigg(\frac{x_1+x_2}2,\,\frac{y_1+y_2}2\bigg)} \end{align} [/tex]
- [tex] \rm Midpoint = \bigg(\frac{\text-1+9}2,\,\frac{\text-4+6}2 \bigg) \\ [/tex]
- [tex] \rm Midpoint = \bigg(\frac{\,8\,}2,\,\frac{\,2\,}2 \bigg) \\ [/tex]
- [tex] \rm Midpoint = (4, \,1) [/tex]
Thus, the center of the circle is at (4, 1). Substitute it to the equation of the circle in standard form that is written as:
- [tex] (x-h)^2 + (y-k)^2 = r^2 [/tex]
Where (h,k) is the center and r is the radius of the circle.
- [tex] (x-4)^2 + (y-1)^2 = r^2 [/tex]
Find the square of the radius by substituting one of the given endpoints of the diameter. We will be using (9,6).
- [tex] (9-4)^2 + (6-1)^2 = r^2 [/tex]
- [tex] (5)^2 + (5)^2 = r^2 [/tex]
- [tex] 25 + 25 = r^2 [/tex]
- [tex] 50 = r^2 [/tex]
Substitute the square of the radius to the given equation with the given center.
- [tex] (x-4)^2 + (y-1)^2 = 50 [/tex]
Therefore, the equation of the circle in standard form is (x-4)² + (y-1)² = 50
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(ノ^_^)ノ
CIRCLE EQUATION
[tex]\huge\bold{✒Solution:}[/tex]
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According to the question, the center of circle [tex] \mathtt{( \frac{ - 1 + 9}{2}, \frac{ - 4 + 6}{2}) \implies \: (4,1)}[/tex] and the diameter:
[tex] \mathtt{\sqrt{ (- 1 - 9)^{2} + ( - 4 - 6)^{2} }}[/tex]
[tex] •\: \mathtt{ \sqrt{100 + 100} }[/tex]
[tex]• \: \mathtt{10 \sqrt{2} }[/tex]
So the radius : [tex] \mathtt{ 5\sqrt{2} }[/tex]
So the equation : [tex] \mathtt{(x - 4) ^{2} + (y - 1)^{2} = ( 5\sqrt{2})^{2} = 50}[/tex]
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[tex]\huge\bold{✒Answer:}[/tex]
[tex]\blue{•••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]
Therefore, the equation of the circle in standard form is (x - 4)² + (y - 1)² = 50
[tex]\blue{•••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]
(ノ^_^)ノ
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