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find an equation of the circle that has center (-5, 0) and passes through (4, -6).

PLEASE HELP !!!!

Find An Equation Of The Circle That Has Center 5 0 And Passes Through 4 6 PLEASE HELP class=

Sagot :

semsee45

Answer:

[tex](x+5)^2+y^2=117[/tex]

Step-by-step explanation:

Equation of a circle

[tex](x-a)^2+(y-b)^2=r^2[/tex]

(where (a, b) is the center and r is the radius)

Given:

  • center = (-5, 0)

[tex]\implies (x-(-5))^2+(y-0)^2=r^2[/tex]

[tex]\implies (x+5)^2+y^2=r^2[/tex]

To find r², input the coordinates of the given point (4, -6) into the equation:

[tex]\implies (4+5)^2+(-6)^2=r^2[/tex]

[tex]\implies 81+36=r^2[/tex]

[tex]\implies r^2=117[/tex]

Therefore, the equation of the circle is:

[tex]\implies (x+5)^2+y^2=117[/tex]

fieryanswererft

Answer:

Standard form = (x + 5)² + y² = 117

General Form = x² + 10x + y² - 92 = 0

Explanation:

(x - h)² + (y - k)² = r²

  • center: (h, k)

Find radius using:

[tex]\sf Distance \ between \ two \ points = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]\rightarrow \sf radius : \sqrt{(-5-4)^+(0-(-6))^2} \ = \ 3\sqrt{13} \ \ units[/tex]

Find equation inserting values:   Given center: (-5, 0)

⇒ (x - (-5))² + (y - 0)² = (3√13)²

⇒ (x + 5)² + y² = 117

⇒ x² + 10x + 25 + y² - 117 = 0

⇒ x² + 10x + y² - 92 = 0

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