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What is the general form of the equation for the given circle centered at O(0, 0)?
a) x2 + y2 + 41 = 0
b) x2 + y2 − 41 = 0
c) x2 + y2 − 41 = 0
d) x2 + y2 + x − y − 41 = 0

What is the general form of the equation for the given circle centered at O0 0 a x2 y2 41 0 b x2 y2 41 0 c x2 y2 41 0 d x2 y2 x y 41 0 class=

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ghanami

 

hello :
an equation of the circle Center at the O(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a =0 and b = 0 (Center at the origin)
r = AO
r² = (AO)²
r² = (4-0)² +(5-0)² = 16+25=41
an equation of the circle that satisfies the stated conditions.
Center at the origin, passing through B(4, 5) is :  x² +y² = 41

x² +y² - 41=0

Equation of a graph is a mathematical statement true only for points of that graph. The equation of specified circle is:Option b) x² + y² −41 = 0

What is the equation of a circle with radius r units, centered at (x,y) ?

If a circle O has radius of r units length and that it has got its center positioned at (h, k) point of the coordinate plane, then, its equation is given as:
[tex](x-h)^2 + (y-k)^2 = r^2[/tex]

What is the distance between two points ( p,q) and (x,y)?

The shortest distance(straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:

D = √[(x-p)² + (y-q)²]    units.

For given case, we have h = 0, and k = 0,

Also, we've got radius = distance between O and B as length of radius, which is

r = √[(4-0)² + (5-0)²] = √[16+25] = √41   units.

Thus, we have the equation as:

(x-h)² + (y-k)² = r²

x²+ y² = (√41  )²

x² + y² = 41

x² + y² - 41 = 0

Thus, The equation of specified circle is:Option b) x² + y² - 41 = 0

Learn more about equation of circle here:

https://brainly.com/question/15848342

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