Question

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.

Rational exponents and radicals
ANSWERED
asked 2020-11-03

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

Expert Answers (1)

2020-11-04

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

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