Question
Let p and q be constants such that the graph of x^2+y^2-6x+py+q=0 is tangent to the y-axis. What is the area of the region enclosed by the graph?
I have no idea what to do? Can someone please help me with the solution to the problem?
Answers
bobpursley
x^2+y^2-6x+py+q=0
(x-3)^2 + (y+p/2)^2=-q-9-P^2/4
so radius of this circle is
sqrt(q-9-p^2/4)
area= PI (q-9-p^2/4)
check my thinking.
(x-3)^2 + (y+p/2)^2=-q-9-P^2/4
so radius of this circle is
sqrt(q-9-p^2/4)
area= PI (q-9-p^2/4)
check my thinking.
Lilly
For the area, wouldn't be (q-9-p^2/4)^2 because the formula for the circle is pi radius squared?