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What is the equation of a circle whose centre is at 3 -5 and passes through the point of 6 -7 on the Cartesian plane?
Wiki User
∙ 7y ago
The equation of a circle with centre (x0, y0) and radius r is given by:
(x - x0)² + (y - y0)² = r²
The circle with centre (3, -5) and passing through the point (6, -7) has radius:
use Pythagoras:
radius = distance from centre (x0, y0) to point (x1, y1) = √((x1 - x0)² + (y1 - y0)²)
→ radius = √((6 - 3)² + (-7 - -5)²) = √(3² + (-2)²) = √(9 + 4) = √13
Thus the equation is:
(x - 3)² + (y - -5)² = (√13)²
→ (x - 3)² + (y + 5)² = 13
Which can be expanded to give:
x² - 6x + 9 + y² +10y + 25 = 13
→ x² - 6x + y² + 10y + 21 = 0
Wiki User
∙ 7y ago
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