phazevfx
Answered

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With the equation of the circle that has a center at the origin and passes through the point (1-6),​

Sagot :

semsee45

Answer:

[tex]\sf x^2+y^2=37[/tex]

Step-by-step explanation:

Equation of a circle:  [tex]\sf(x-h)^2+(y-k)^2=r^2[/tex]

where (h, k) is the center and r is the radius

Given the center is at (0, 0)

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

[tex]\sf \implies x^2+y^2=r^2[/tex]

Given the circle passes through point (1, -6):

[tex]\sf \implies (1)^2+(-6)^2=r^2[/tex]

[tex]\sf \implies r^2=37[/tex]

Therefore, the equation of the circle is:

[tex]\sf \implies x^2+y^2=37[/tex]

Equation of a circle passing through origin

  • x²+y²=r²

So

As (1,-6) lies on circle it will satisfy

  • 1²+(-6)²=r^2
  • 1+36=r²
  • r²=37

Now

equation of the circle

  • x²+y²=37
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