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What is the center of the circle x^2 + y^2 + 3x + 4y = 45

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lxsptter | Student, Undergraduate | Valedictorian

Posted May 29, 2013 at 6:01 PM via web

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What is the center of the circle x^2 + y^2 + 3x + 4y = 45

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 29, 2013 at 6:04 PM (Answer #1)

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

Express the equation of the circle given in the form described.

x^2 + y^2 + 3x + 4y = 45

=> x^2 + 3x + y^2 + 4y = 45

=> x^2 + 3x + (3/2)^2 + y^2 + 4y + 4 = 45 + 9/4 + 4

=> (x + 3/2)^2 + (y + 2)^2 = 205/4

The center of the circle is (-3/2, -2)

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oldnick | Valedictorian

Posted May 30, 2013 at 1:36 AM (Answer #2)

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`A)`  You can re-write the eqaution as:

`[x-2]^2+[x-(-3)]^2=5^2`

So the circle has the centre in the point C(2;-3)

`B)`   to se if the point A(6-6) lies on the cirlce is enough to see if the distance  d from the center C(2,-3) is R.

indeed :     

 `d^2=(6-2)^2+(-6+3)^2=4^2+3^2=5^2=R`

The the point lies on the circle.

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