Answer :
Answer:
The radius of [tex]x^{2}+y^{2}-18\cdot x - 14\cdot y +124 = 0[/tex] is [tex]\sqrt{6}[/tex].
Step-by-step explanation:
The given expression must be converted into the standard form of a circle, which is described by:
[tex](x-h)^{2}+(y-k)^{2} = r^{2}[/tex]
Where:
[tex]h[/tex], [tex]k[/tex] - Location of center, dimensionless.
[tex]r[/tex] - Radius, dimensionless.
The expression is modified algebraically as follows:
1) [tex]x^{2}+y^{2}-18\cdot x - 14\cdot y +124 = 0[/tex] Given
2) [tex](x^{2}-18\cdot x)+(y^{2}-14\cdot y)+124 = 0[/tex] Commutative and associative properties.
3) [tex](x^{2}-18\cdot x+81)+(y^{2}-14\cdot y +49)+124 = 81+49[/tex] Commutative and associative properties/Compatibility with addition
4) [tex](x-9)^{2}+(y-7)^{2} + 124 = 130[/tex] Perfect square binomial/Definition of sum
5) [tex](x-9)^{2}+(y-7)^{2} = 6[/tex] Compatibility with addition/Existence of additive inverse/Modulative property/Result
Given that [tex]r^{2} = 6[/tex], the radius of the circle is [tex]\sqrt{6}[/tex].