What is the center of the circle represented by y^2-16y+x^2-8x-176 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The center of the circle represented by y^2-16y+x^2-8x-176 = 0 has to be determined. The equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2. Express the given relation in a similar format.

y^2 - 16y + x^2 - 8x - 176 = 0

=> y^2 - 2*8*y + 64 + x^2 - 2*4*x + 16 - 176 - 64 - 16 = 0

=> (y - 8)^2 + (x - 4)^2 = 256

This gives the center of the circle as (4, 8) and the radius is 16.

The center of the circle represented by y^2-16y+x^2-8x-176 = 0 is (4, 8)

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jgor | Student, Grade 10 | eNotes Newbie

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y^2 - 16y + x^2 - 8x = 176

(y^2 - 16y + 64) + (x^2 - 8x + 16) = 176 + 64 + 16

(y-8)^2 + (x-4)^2 = 256

the zeros are 8 and 4 so the center is (4,8) corresponding to the x and y in the equation

the radius is the square root of 256 which is 16

answer: center: (4,8) radius: 16

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