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Write the standard equation of the circle with the center ​(-14​,-5​) that passes through the point ​(​-7,​5).

Respuesta :

fieryanswererft

equation: (x + 14)² + (y + 5)² = 149

Given:

  • centre :  ​(-14​,-5​)
  • point ​(​-7,​5)

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Formula's:

  • (x-h)² + (y-k)² = r²
  • centre : (h, k)
  • radius : r
  • distance between points : [tex]\sf \sqrt{(x2-x1)^2 + (y2-y1)^2}[/tex]

Find the radius:

[tex]\rightarrow \sf \sqrt{(-7-(-14))^2 + (5-(-5))^2}[/tex]

[tex]\sf \rightarrow \sqrt{\left(-7+14\right)^2+\left(5+5\right)^2}[/tex]

[tex]\sf \rightarrow \sqrt{149}[/tex]

Equation of circle:

  • (x-h)² + (y-k)² = r²
  • (x-(-14))² + (y-(-5))² = (√149)²
  • (x + 14)² + (y + 5)² = 149

Graph for clarification:

Ver imagen fieryanswererft
semsee45

Answer:

[tex]\sf (x+14)^2+(y+5)^2=149[/tex]

Step-by-step explanation:

Standard equation of a circle:  [tex]\sf (x-a)^2+(y-b)^2=r^2[/tex]

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

Substitute the given center (-14, -5) into the equation:

[tex]\sf \implies (x-(-14))^2+(y-(-5))^2=r^2[/tex]

[tex]\sf \implies (x+14)^2+(y+5)^2=r^2[/tex]

Now substitute the point (-7, 5) into the equation to find r²:

[tex]\sf \implies ((-7)+14)^2+(5+5)^2=r^2[/tex]

[tex]\sf \implies (7)^2+(10)^2=r^2[/tex]

[tex]\sf \implies 149=r^2[/tex]

Final equation:

[tex]\sf (x+14)^2+(y+5)^2=149[/tex]

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