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What is the centre and radius of a circle on the Cartesian plane whose equation is 100x squared plus 100y squared -120x plus 100y -39 equals 0?
Wiki User
∙ 8y ago
A circle with centre (x0, y0) and radius r has the formula:
(x - x0)² + (y - y0)² = r²
→ x² - 2x0 x + x0² + y² - 2y0 y + y0² - r² = 0
Thus (by completing the square):
100x² + 100y² - 120x + 100y - 39 = 0
→ x² + y² - 1.2x + y - 0.39 = 0
→ x² - (2 × 0.6)x + 0.6² - 0.6² + y² - (2 × 0.5)y + 0.5² - 0.5² - 0.39 = 0
→ (x - 0.6)² + (y + 0.5)² - 0.36 - 0.25 - 0.39 = 0
→ (x - 0.6)² + ( y - -0.5)² = 1 = 1²
→ the circle has centre (0.6, -0.5) and radius of 1.
Another Answer:-
Equation: 100x^2+100y^2 -120x+100y -39 = 0
Divide all terms by 100: x^2+y^2 -6/5x+y -39/100
Completing the squares of x and y: (x-3/5)^2+(y+1/2)^2 -9/25 -1/4 -39/100 = 0
So: (x-3/5)^2+(y+1/2)^2 = 1
Therefore the centre of the circle is at (3/5, -1/2) and its radius is 1
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∙ 8y ago
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