Question 1028942: Decide whether or not the equation has a circle as its graph. If it does, give the center
and the radius. If it does not, describe the graph.
x^2 + y^2 - 4x - 10y + 29 = 0
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
Decide whether or not the equation has a circle as its graph. If it does, give the center
and the radius. If it does not, describe the graph.
x^2 + y^2 - 4x - 10y + 29 = 0
Solution:
This is a simpler version of question 1028943:
https://www.algebra.com/tutors/faq.mpl?key=1028943;action=answer_question
because the coefficients of x² and y² are already unity (=1).
As explained in the above link, the equation represents a circle.
Recall that the general equation of a circle is given by:
(x-a)²+(y-b)²=r² ............ (1)
The circle is centred at (a,b) with radius r.
We'll need to complete squares of the given equation and compare with (1) to find the centre (a,b) and radius r.
x^2 + y^2 - 4x - 10y + 29 = 0 [prepare for completing squares]
=>
x²-2(2x) + y²-2(5y) = -29 [ complete squares ]
=>
(x²-2(2x)+4) + (y²-2(5y)+5²) = -29+4+25 [factor left side, and simplify right-side]
=>
(x-2)²+(y-5)²=0.............(2) [ equation of a circle ]
=>
Comparing (2) with equation (1), we can conclude:
a=2,
b=5,
r=0
That means that we have a circle centred at (2,5) with radius zero.
This is a degenerate circle which degenerates to a point, but the equation still represents a circle, even though only a point is visible on the graph.
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