________________________________, Note: Please Answer My Question Correctly, Nonsense Answer Will Be Reported. ________________________________ , Lesson: Illustrating The Center-Radius Of The Equation Of A Circle ________________________________

________________________________

Note: Please answer my question correctly, nonsense answer will be reported. ________________________________

Lesson: Illustrating the Center-Radius of the Equation of a Circle ________________________________

✒️CIRCLE EQUATIONS

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$\largenderline{\mathbb{ANSWER}:}$

Part A:

$\qquad \large \rm 1) \; (x - 2)^2 + (y + 4)^2 = 25$

$\qquad \large \rm 2) \; (x - 3)^2 + y^2 = 1/9$

$\qquad \large \rm 3) \; x^2 + y^2 = 1/16$

$\qquad \large \rm 4) \; (x + 7)^2 + (y - 1)^2 = 5$

$\qquad \large \rm 5) \; (x + 6)^2 + (y + 3)^2 = 48$

$\qquad \large \rm 6) \; x^2 + (y - 6)^2 = 81$

Part B:

$\qquad \large \rm 7) \; (x - 3)^2 + (y + 4)^2 = 45$

$\qquad \large \rm 8) \; (x - 5)^2 + (y - 1)^2 = 18$

$\qquad \large \rm 9) \; x^2 + y^2 = 25$

$\qquad \large \rm 10) \; x^2 + y^2 = 16$

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$\largenderline{\mathbb{SOLUTION}:}$

For Part A and B: The equation of the circle in standard form is written as:

$(x - h)^2 + (y - k)^2 = r^2$

Where (h,k) is the center and r is the radius. Substitute each given to get its equation.

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“Part A”

Number 1:

$(x - 2)^2 + \bigy - (\text-4)\big^2 = 5^2$

$(x-2)^2 + (y + 4)^2 = 25$

Number 2:

$(x - 3)^2 + (y - 0)^2 = (1/3)^2$

$(x - 3)^2 + y^2 = 1/9$

Number 3:

$(x - 0)^2 + (y - 0)^2 = (1/4)^2$

$x^2 + y^2 = 1/16$

Number 4:

$\bigx - (\text-7)\big^2 + (y - 1)^2 = (\sqrt5\,)^2$

$(x + 7)^2 + (y - 1)^2 = 5$

Number 5:

$\bigx - (\text-6)\big^2 + \bigy - (\text-3)\big^2 = (4\sqrt3\,)^2$

$(x + 6)^2 + (y + 3)^2 = 16(3)$

$(x + 6)^2 + (y + 3)^2 = 48$

Number 6:

$(x - 0)^2 + (y -6)^2 = 9^2$

$x^2 + (y - 6)^2 = 81$

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“Part B”

Number 7:

Substitute the given center in the standard form of the equation.

$(x - 3)^2 + \bigy - (\text-4)\big^2 = r^2$

$(x - 3)^2 + (y + 4)^2 = r^2$

Find the square of the radius if it passes through (6,2)

$(6 - 3)^2 + (2 + 4)^2 = r^2$

$(3)^2 + (6)^2 = r^2$

$9 + 36 = r^2$

$45 = r^2$

Substitute the square of the radius to the equation.

$(x - 3)^2 + (y + 4)^2 = 45$

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Number 8:

Substitute the given center in the standard form of the equation.

$(x - 5)^2 + (y - 1)^2 = r^2$

Find the square of the radius if it passes through (8,-2)

$(8 - 5)^2 + (\text-2 - 1)^2 = r^2$

$(3)^2 + (\text-3)^2 = r^2$

$9 + 9 = r^2$

$18 = r^2$

Substitute the square of the radius to the equation.

$(x - 5)^2 + (y - 1)^2 = 18$

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Number 9:

Substitute (0,0) as the given center in the standard form of the equation since it is at the origin.

$(x - 0)^2 + (y - 0)^2 = r^2$

$x^2 + y^2 = r^2$

Find the square of the radius if it passes through (4,3)

$4^2 + 3^2 = r^2$

$16 + 9 = r^2$

$25 = r^2$

Substitute the square of the radius to the equation.

$x^2 + y^2 = 25$

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Number 10:

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

$\begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm Midpoint = \bigg(\frac{x_2+x_1}2,\,\frac{y_2+y_1}2\bigg)} \end{align}$

$\rm Center = \bigg(\frac{\text-4 + 4}2,\, \frac{0+0}2 \bigg) \\$

$\rm Center = \bigg(\frac{\,0\,}2,\, \frac{\,0\,}2 \bigg) \\$

$\rm Center = (0,0)$

Thus, the center is at the origin. Substitute it in the standard form of the equation.

$(x - 0)^2 + (y - 0)^2 = r^2$

$x^2 + y^2 = r^2$

Find the square of the radius if it passes through one of the given endpoints of the diameter: (4,0)

$4^2 + 0^2 = r^2$

$16 + 0 = r^2$

$16 = r^2$

Substitute the square of the radius to the equation.

$x^2 + y^2 = 16$

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(ノ^_^)ノ

, Note: Please answer my question correctly, nonsense answer will be reported. , Lesson: Illustrating the Center-Radius of the Equation of a Circle