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radius = ½ diameter

= ½ Distance between endpoints

= ½√((difference in x)² + (difference in y)²)

= ½√((8 - 2)² + (7 - -3)²)

= ½√(6² + 10²)

= ½√136

Centre of circle = midpoint of diameter

= (mean average of x, mean average of y)

= ((2 + 8)/2, (-3 + 7)/2)

= (5, 2)

Circle with centre (Xc, Yc) and radius r has equation:

(x - Xc)² + (y - Yc)² = r²

Which for circle with centre (5, 2) and radius ½√136 gives:

(x - 5)² + (y - 2)² = (½√136)² = ¼ × 136 = 34

→ (x - 5)² + (y - 2)² = 34

This can be expanded and rearranged to get:

x² -10x + 25 + y² - 4y + 4 - 34 = 0

→ x² - 10x + y² - 4y - 5 = 0

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Q: What is the equation of a circle when its diameter endpoints are at 2 -3 and 8 7?
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