✒️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]
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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