Find the center of the circle x^2 + 3x + y^2 - 8y + 4 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2

x^2 + 3x + y^2 - 8y + 4 = 0

Use completion of squares

=> x^2 + 3x + 9/4 + y^2 - 8y + 16 + 4 - 9/4 - 16 = 0

=> (x + 3/2)^2 + (y - 4)^2 = 14.25

This gives the center of the circle x^2 + 3x + y^2 - 8y + 4 = 0 as (-3/2, 4)

heba1986's profile pic

heba1986 | Teacher | (Level 1) eNoter

Posted on

if we want to know the center of the circle

x^2+3x+y^2-8y+4=0

we should change the equation of circle to the general form

(x-a)^2+(y-b)^2=r^2

this equation for circle the center for it is (a,b)and the radious is r

so we use completion of squares:

x^2+3x+(9/4)-(9/4)+y^2-8y+16-16+4=0

(x+3/2)^2+(y-4)^2=14.25

=>the center for the circle is (-3/2,4)

 

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