Question

A. 1

B. 7
54. What is the center and radius of the circle given by
8x2 + 8y2- 16x - 32y + 24 = 0?
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Answer 1

Explanation:

Given equation of circle (in general form) is:

`8 {x}^(2) + 8 {y}^(2) - 16x - 32y + 24 = 0 \\ \\ dividing \: throughout \: by \: 8 \: we \: find : \\ {x}^(2) + {y}^(2) - 2x - 4y + 3 = 0 \\ equating \: it \: with : \\ {x}^(2) + {y}^(2) + 2gx + 2fy + c = 0 \\ 2g = - 2 \implies \: g = - 1\\ 2f = - 4 \implies \: f = - 2 \\ c = 3 \\ \therefore \:center = ( - g, \: \: - f) = (1, \: \: 2) \\ \\ radius \: r = \sqrt{ {g}^(2) + {f}^(2) - c } \\ = \sqrt{ {1}^(2) + {2}^(2) - 3} \\ = √(1 + 4 - 3) \\ = √(5 - 3) \\ = √(2) \: units`