# Find the equation of the circle whose center (0,13) and the area = 25pi.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given the center of a circle is the point (0, 13). Also, given that the area of the circle = 25*pi.

We will write the equation of the circle.

(x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius.

==> (x-0)^2 + (y-13)^2 = r^2

==> x^2 + ( y-13)^2 = r^2.

Now we need to determine the radius.

Given the area of the circle = 25pi.

==> r^2 * pi = 25*pi

==> r^2 = 25

==> r= 5

Then the radius of the circle = 5

==> x^2 + ( y-13)^2 = 5^2

==> x^2 + ( y-13)^2 = 25

==> x^2 + y^2 - 26y + 144 = 0

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write the equation of the circle in standard form:

(x-h)^2 + (y-k)^2 = r^2, where h and k are the coordinates of the center of the circle.

We'll identify h and k:

h = 0 and k = 13

Now, we'll determine the radius of the circle, using the formula for area of the circle:

A = pi*r^2

25*pi = pi*r^2

We'll divide by pi:

r^2 = 25

r = 5

We'll accept only the positive value, since it is about a radius of a circle and it cannot be negative.

Now,we'll substitute the coordinates of the center of the circle and the value of radius in the equation of the circle:

x^2 + (y - 13)^2 = 25

If we'll expand the square, we'll obtain the general form of the equation:

x^2 + y^2 - 26y + 169 - 25 = 0

x^2 + y^2 - 26y  + 144 = 0

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The area of the circle with radius r is pir^2.

Therefore  if the actual area  is 25pi, then pir^2 = pi*25 => r^2= 25 => r= 5.

So the radius r = 5 and the centre is C(0,13)

The equation of the circle with centre C(h,k) and radius r is (x-h)62+(y-k)^2 = r^2.

Therefore the equation of the circle with C(0,13) and r = 5 is

(x-0)^2+(y-13)^2 = 5^2.

x^2+y^2-26y+169 = 25.

x^2+y^2 -26y - 144 = 0 is the equation if the circle.