- #1
odolwa99
- 85
- 0
I'm having some difficulty with this question. Can anyone help me out, please?
Many thanks.
A circle of radius length [itex]\sqrt{20}[/itex] contains the point (-1, 3). Its centre lies on the line x + y = 0. Find the equations of the 2 circles that satisfy these conditions.
Circle equation for the point: [itex](-1-h)^2+(3-k)^2=20\rightarrow h^2+k^2+2h-6k-10=0[/itex]
Because x+y=0 contains centre c as (-g, -f), we can say -g - f = 0.
Thus, -g = -2h & -f = 6k.
And so, [itex]x+y=0\rightarrow -2h+6k=0\rightarrow h=3k[/itex].
Returning to circle equation: [itex](3k)^2+k^2+2(3k)-6k-10=0\rightarrow 10k^2=10\rightarrow k=+/- 1[/itex]
For k = 1: [itex]h=3k\rightarrow h=3[/itex].
Thus, centre c = (3, 1) and equation = [itex](x-3)^2+(y-1)^2=20\rightarrow x^2+y^2-6x-2y-10=0[/itex]
For k = -1: [itex]h=-3[/itex]
Thus, centre c = (-3, -1) and equation = [itex](x+3)^2+(y+1)^2=20\rightarrow x^2+y^2+6x+2y-10=0[/itex]
Ans.: (From textbook): [itex]x^2+y^2-2x+2y-18=0[/itex] & [itex]x^2+y^2+10x-10y+30=0[/itex]
Many thanks.
Homework Statement
A circle of radius length [itex]\sqrt{20}[/itex] contains the point (-1, 3). Its centre lies on the line x + y = 0. Find the equations of the 2 circles that satisfy these conditions.
Homework Equations
The Attempt at a Solution
Circle equation for the point: [itex](-1-h)^2+(3-k)^2=20\rightarrow h^2+k^2+2h-6k-10=0[/itex]
Because x+y=0 contains centre c as (-g, -f), we can say -g - f = 0.
Thus, -g = -2h & -f = 6k.
And so, [itex]x+y=0\rightarrow -2h+6k=0\rightarrow h=3k[/itex].
Returning to circle equation: [itex](3k)^2+k^2+2(3k)-6k-10=0\rightarrow 10k^2=10\rightarrow k=+/- 1[/itex]
For k = 1: [itex]h=3k\rightarrow h=3[/itex].
Thus, centre c = (3, 1) and equation = [itex](x-3)^2+(y-1)^2=20\rightarrow x^2+y^2-6x-2y-10=0[/itex]
For k = -1: [itex]h=-3[/itex]
Thus, centre c = (-3, -1) and equation = [itex](x+3)^2+(y+1)^2=20\rightarrow x^2+y^2+6x+2y-10=0[/itex]
Ans.: (From textbook): [itex]x^2+y^2-2x+2y-18=0[/itex] & [itex]x^2+y^2+10x-10y+30=0[/itex]