Question 1028943: Decide whether or not the equation below has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph.
49x^2 + 49y^2 - 42x + 42y - 63 = 0
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
Decide whether or not the equation below has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph.
49x^2 + 49y^2 - 42x + 42y - 63 = 0
Solution:
A second-degree equation represents a circle if both of the following conditions are met:
1. both the x² and y² terms are present, and
2. the coefficients of both terms are positive AND equal.
Since the given equation has the same positive coefficient of +49 for both the x² and y² terms, the equation represents a circle.
The general equation of a circle centred at (a,b) and radius r is given by:
(x-a)²+(y-b)²=r² ............ (1)
To find the centre and radius of the given circle, we divide by the common coefficient and complete squares as follows, then compare the resulting equation with equation (1) above to find a, b and r.
49x^2 + 49y^2 - 42x + 42y - 63 = 0 [divide by 49]
=>
x² + y² -2(21/49)x + 2(21/49)y - 63/49 = 0 [simplification and exchange terms to prepare for completing squares]
=>
x²-2(3/7)x + y² + 2(3/7)y = 9/7 [ add terms on both sides to complete squares]
=>
(x²-2(3/7)x+(3/7)²) + (y²+2(3/7)y+(3/7)²) = 9/7+2(3/7)² [using identity a²+2ab+b²≡(a+b)²]
=>
(x-3/7)² + (y+3/7)² = (9/7)² ............(2)
Comparing (2) with (1), we can deduce that:
a=3/7
b=-3/7
r=9/7
Hence the circle has centre at (3/7, -3/7) and radius 9/7.
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