What is the center of the circle represented by the equation: x^2 + y^2 – 8x – 4y + 4 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We know that a circle with radius r and with the coordinates of its center being (a, b) can be represented as (x - a) ^2 + (y - b) ^2 = r^2

Now the equation of the circle we have been given is x^2+ y^2 – 8x– 4y + 4 = 0

Let us convert this to the standard form

x^2+ y^2 – 8x– 4y + 4 = 0

=> x^2 – 8x + y^2 – 4y + 4 =0

Add 16 to the complete the square with the x term and add 4 to complete the square with the y term. We also have to subtract 20 for the 16 and 4 we have added.

=> x^2 – 8x + 16 + y^2 – 4y + 4 + 4 - 20 =0

=> (x – 4) ^2 + (y -2) ^2 – 16 = 0

=> (x – 4) ^2 + (y -2) ^2 = 16

=> (x – 4) ^2 + (y -2) ^2 = 4^2

The equation is now in the form of the standard equation of a circle with a= 4, b= 2 and r = 4.

Therefore the coordinates of the center of the circle are (4, 2)

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neela | High School Teacher | (Level 3) Valedictorian

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We are required to find the centre of the circle whose equation is  x^2 + y^2 – 8x – 4y + 4 = 0.

Solution:

If  (h , k ) is the centre of a circle with radius r, then any point (x,y) on the circumference of the circle is at a distance r from the centre (h,k).

Therefore  by distance formula, r^2 = (x-h)^2+(y-k)^2.

Expanding the right side, we get:

Or r^2 = x^2 -2hx+h^2+ y^2-2yk +k^2.

Or x^2+y^2 -2hx- 2ky +(h^2+k^2-r^2) .....(1) is the equation of the circle with centre (h, k) and radius r.

The given equation is :

 x^2 + y^2 – 8x – 4y + 4 = 0 x^2 + y^2 – 8x – 4y + 4 = 0.....(2)

Equating the coefficients of x , y and constant terms, we get:

 - 2h = -8 , or h = -8/-2 = 4.

-2k = -4 , or k = -4/-2 = 2.

Therefore (h, k ) = (4, 2). So  (4,2) are the coordinates of the centre of the given circle.

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