How do you evaluate #(x-6)^{2}+(y+5)^{2}=15^{2}#?

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Answer 1 · 2017-09-20T13:18:55

See explanation

#color(blue)("The teaching bit")#

This is the equation of a circle. Also note that it is of the form of a Pythagorean triangle #->x^2+y^2=r^2#

The changes to #x and y# indicate the circles centre. Multiply each value of change by #(-1)# and you have the centres coordinates.

#color(blue)("Answering the question")#

Given: #(x-6)^2+(y+5)^2=15^2#

Assumption: you are required to make #y# the 'dependant 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+12x-36# giving:

#(y+5)^2=-x^2+12x+189#

Square root both sides

#y+5=sqrt(-x^2+12x+189)#

#y=sqrt(-x^2+12x+189)-5#

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