Answered

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given the center and the radius of the circle, write the standard form and general form of the equation of the circle​

Given The Center And The Radius Of The Circle Write The Standard Form And General Form Of The Equation Of The Circle class=

Sagot :

KAntoineDoixidn

✒️CIRCLE EQUATIONS

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

Standard Form:

[tex] \quad \large \rm 1) \; (x-4)^2 + (y-5)^2 = 4 [/tex]

[tex] \quad \large \rm 2) \; (x-3)^2 + (y+2)^2 = 9 [/tex]

General Form:

[tex] \quad \large \rm 1) \; x² + y² - 8x - 10y + 37 = 0 [/tex]

[tex] \quad \large \rm 2) \; x² + y² - 6x + 4y + 4 = 0 [/tex]

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

First, write 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. After that, rearrange it into general form that is written as:

  • [tex] x^2 + y^2 + Ax + By + C = 0 [/tex]

Substitute the given center and radius to determine its equations.

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Number 1:

Substitute (4,5) as the center and 2 as the radius to determine the equation in standard form.

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

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

Therefore, the equation of the given circle in standard form is (x-4)² + (y-5)² = 4. Expand and rearrange the terms to get the general form.

  • [tex] \small (x-4)^2 + (y-5)^2 = 4 [/tex]

  • [tex] \small x^2 - 8x + 16 + y^2 - 10y + 25 = 4 [/tex]

  • [tex] \small x^2 + y^2 - 8x - 10y + 16 + 25 - 4 = 0 [/tex]

  • [tex] \small x^2 + y^2 - 8x - 10y + 41 - 4 = 0 [/tex]

  • [tex] \small x^2 + y^2 - 8x - 10y + 37 = 0 [/tex]

Therefore, the equation of the given circle in general form is x² + y² - 8x - 10y + 37 = 0

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Number 1:

Substitute (4,5) as the center and 2 as the radius to determine the equation in standard form.

  • [tex] (x-3)^2 + \big[y-(\text-2)\big]^2 = 3^2 [/tex]

  • [tex] (x-3)^2 + (y+2)^2 = 9 [/tex]

Therefore, the equation of the given circle in standard form is (x-3)² + (y+2)² = 9. Expand and rearrange the terms to get the general form.

  • [tex] \small (x-3)^2 + (y+2)^2 = 9 [/tex]

  • [tex] \small x^2 - 6x + 9 + y^2 + 4y + 4 = 9 [/tex]

  • [tex] \small x^2 + y^2 - 6x + 4y + 9 + 4 - 9 = 0 [/tex]

  • [tex] \small x^2 + y^2 - 6x + 4y + 13 - 9 = 0 [/tex]

  • [tex] \small x^2 + y^2 - 6x + 4y + 4 = 0 [/tex]

Therefore, the equation of the given circle in general form is x² + y² - 6x + 4y + 4 = 0

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