Question

what is the radius of the circle shown below?
x^2+y^2-8x-10y-8=0

Answer 1

complete the square in both x and y

Answer 2

\[\left\{ x^2-8x+\left( \frac{ -8 }{ 2 } \right)^2 \right\}+\left\{ y^2-10 y+\left( \frac{ -10}{ 2 } \right) ^2 \right\}=8+\left( \frac{ -8 }{ 2 } \right)^2+\left( \frac{ -10 }{ 2 } \right)^2\]
?

Answer 3

what do i do with the -8?

Answer 4

sorry i just dont which one the -8 belongs too or if i have to add it to the other side

Answer 5

\[\left( x-h \right)^2+\left( y-k \right)^2=r^2\]

Answer 6

add it to both sides of the equation so you have terms in x and y on the left... and a value on the right

Answer 7

so to start you have

x^2 - 8x + y^2 - 10y = 8

Answer 8

(x^2-8x)(y^-10y)=8

Answer 9

thats the start now complete the squares... and remember to add the same values to both sides of the equation.

Answer 10

i got (4,5) as the center

Answer 11

thats correct..

Answer 12

so what is the radius..?

Answer 13

7

Answer 14

thats correct * + 25 + 16 = 49