Respuesta :

[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-6)^2+(y+5)^2=15^2\implies [x-6]^2+[y-(-5)]^2=15^2~~~~ \begin{cases} \stackrel{radius}{15}\\\\ \stackrel{center}{(6,-5)} \end{cases}[/tex]

demolisher

Answer:

5

Step-by-step explanation

Given: ( x − 6 ) 2 + ( y + 5 ) 2 = 15 2 Assumption: you are required to make y the 'dependent variable' (the answer).

Subtract ( x − 6 ) 2 from both sides ( y + 5 ) 2 = 15 2 − ( x − 6 ) 2

Expanding the − ( x − 6 ) 2 → − x 2 + 12 x − 36 giving: ( y + 5 ) 2 = − x 2 + 12 x + 189

Square root both sides y + 5 = √ − x 2 + 12 x + 189 y = √ − x 2 + 12 x + 189 − 5

Otras preguntas