frenzal_dude
UTS Student
Hi, I had a bit of trouble with this question:
The point (2,1) lies on the circle with equation x^2 + y^2 + 6x - 2y -15 = 0
Find the coordinates of the other end of the diameter through (2,1)
I worked out the centre of the circle to be (-3,1) and radius = 5.
So I tried to use the distance formula to work out the distance between (2,1) and (x,y) and then make x the subject and sub that back into the formula for the circle, however I ended up getting a square root with a negative inside.
btw the answer is: (-8,1)
Thanks for the help.
The point (2,1) lies on the circle with equation x^2 + y^2 + 6x - 2y -15 = 0
Find the coordinates of the other end of the diameter through (2,1)
I worked out the centre of the circle to be (-3,1) and radius = 5.
So I tried to use the distance formula to work out the distance between (2,1) and (x,y) and then make x the subject and sub that back into the formula for the circle, however I ended up getting a square root with a negative inside.
btw the answer is: (-8,1)
Thanks for the help.