geometry problem?

Question

At what point does the circle with the equation x^2+y^2-16x-16y-41=0?
please list out the whole process, I got the binomial as (x-8)^2+(y-8)^2=169 then I set the y equals 0 and solve for x, but the answer is quite different from the key!!!! please give me some idea how the heck do I get the answer of x=5 and x=-21??

Answer 1

(a) You must have LEFT OUT a crucial part of the problem: at what point(s) does this circle do WHAT ?! ; and

(b) I suspect that you'll find it should have been ' + 16 x,' NOT
' - 16 x.' (!!)

Assuming (b), the equation for the circle then becomes:

(x + 8)^2 + (y - 8)^2 = 169 = 13^2,

i.e. a circle centred on (-8, +8), with radius 13.

In that case, the two x-extremities are at (- 21, 8) and (+5, 8).

Thus, if the missing part (a) of the question completed it as follows:

(a) At which points does the circle x^2+y^2+16x-16y-41=0 intersect the line y = 8 ? ,

you'd then have your answer !

PLEASE do get back to us and tell us whether my Vulcan mind-melding technique has divined not only what the answer is, but also WHAT THE QUESTION WAS !

Live long and prosper.

Answer 2

the answer is 2....like 2 points