Solve the equation for this circle in StandardThe form:x^{2} - 16x + y^{2} - 10y + 73 = 0Enter radicals such as sqrt{15} as "sqrt{15}"instead of using rational exponents.

UkusakazaL

UkusakazaL

Answered question

2020-11-03

Solve the equation for this circle in Standard
The form:
x2  16x + y2  10y + 73=0
Enter radicals such as 15 instead of using rational exponents.

Answer & Explanation

tabuordy

tabuordy

Skilled2020-11-04Added 90 answers

Consider the equation of circle x2  16x + y2  10y + 73=0
To change the equation into standard form of the equation (x  h)2 + (y  k)2=r2
Grouping the terms
(x2  16x) + (y2  10y) + 73=0
Add and subtract 64 into (x2 16x)
and Add and subtract 25 into (y2 10y)
(x2 16x + 64  64) + (y2  10y + 25  25) + 73=0
(x2  16x + 64) + (y2 10y + 25)  25  64 + 73=0
(x  8)2 + (y  5)2 16=0
(x  8)2 + (y  5)2=16
Further simplify it
(x  8)2 + (y  5)2=42
Which is standard form of the circle.
Compare it by standard form of the equation (x  h)2 + (y  k)2=r2
h=8, k=5 and r=4
Therefore, centre (h, k)=(8, 5)
And radius r=4
List four point on thw circle (x  8)2 + (y  5)2=42 is given below
(8, 1), (12, 5), (8, 9) and (4, 5)

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